OP Malhotra Class 10 Maths Solutions Chapter 6 Ratio and Proportion Ex 6(c)

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S Chand Class 10 ICSE Maths Solutions Chapter 6 Ratio and Proportion Ex 6(c)

Question 1.
If \(\frac { x }{ a }\) = \(\frac { y }{ b }\) = \(\frac { z }{ c }\), show that
(i) \(\frac{x^3}{a^3}-\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{x y z}{a b c}\)
(ii) \(\left(\frac{a^2 x^2+b^2 y^2+c^2 z^2}{a^3 x+b^3 y+c^3 z}\right)^{\frac{3}{2}}=\sqrt{\frac{x y z}{a b c}}\)
(iii) \(\frac{x z+a c}{x z-a c}=\frac{y z+b c}{y z-b c}\)
(iv) abc\(\left(\frac{x+a}{a}+\frac{y+b}{b}+\frac{z+c}{c}\right)^3\) = 27(x + a) (y + b) (z + c)
(v) each ratio is equal to \(\left(\frac{3 x^3+5 y^3+7 z^3}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}}\)
Solution:
∵ \(\frac { x }{ a }\) = \(\frac { y }{ b }\) = \(\frac { z }{ c }\) (suppose)
∴ x = ak, y = bk, z = ck
(i) \(\frac{x^3}{a^3}-\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{x y z}{a b c}\)
L.H.S. = \(\frac{a^3 k^3}{a^3}-\frac{b^3 k^3}{b^3}+\frac{c^3 k^3}{c^3}\)
= k³ – k³ + k³ = k³
R.H.S = \(\frac{x y z}{a b c}=\frac{a k \cdot b k \cdot c k}{a b c}\)
∴ L.H.S = R.H.S

(ii) \(\left(\frac{a^2 x^2+b^2 y^2+c^2 z^2}{a^3 x+b^3 y+c^3 z}\right)^{\frac{3}{2}}=\sqrt{\frac{x y z}{a b c}}\)

(iii) \(\frac{x z+a c}{x z-a c}=\frac{y z+b c}{y z-b c}\)

(iv) abc\(\left(\frac{x+a}{a}+\frac{y+b}{b}+\frac{z+c}{c}\right)^3\) = 27(x + a) (y + b) (z + c)

(v) Each ratio is equal to \(\left(\frac{3 x^3+5 y^3+7 z^3}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}}\)

Hence proved.

Question 2.
If \(\frac { a }{ b }\) = \(\frac { c }{ d }\) = \(\frac { e }{ f }\), prove the following:
(i) \(\frac{p a^3+q c^3+r e^3}{p b^3+q d^3+r f^3}=\frac{a c e}{b d f}\)
(ii) \(\sqrt{\frac{a^4+c^4}{b^4+d^4}}=\frac{p a^2+q c^2}{p b^2+q d^2}\)
(iii) \(\frac{2 a^4 b^2+3 a^2 e^2-5 e^4 f}{2 b^6+3 b^2 f^2-5 f^5}=\frac{a^4}{b^4}\)
Solution:
\(\frac { a }{ b }\) = \(\frac { c }{ d }\) = \(\frac { e }{ f }\) = k
a = bk, c = dk, e = fk
(i) \(\frac{p a^3+q c^3+r e^3}{p b^3+q d^3+r f^3}=\frac{a c e}{b d f}\)

(ii) \(\sqrt{\frac{a^4+c^4}{b^4+d^4}}=\frac{p a^2+q c^2}{p b^2+q d^2}\)

(iii) \(\frac{2 a^4 b^2+3 a^2 e^2-5 e^4 f}{2 b^6+3 b^2 f^2-5 f^5}=\frac{a^4}{b^4}\)

Question 3.
If a,b,c are in continued proportion, prove that :
(i) (a + b + c) (a – b + c) = a² + b² + c²
(ii) \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
(iii) \(\frac{a^3+b^3+c^3}{a^2 b^2 c^2}=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)
(iv) (4a² + lab + 9b²) : (4b² + 7bc + 9c²) = a : c
Solution:
∵ a, b, c are in continued proportion
Then \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k (Suppose)
∴ b = ck, a = bk = ck.k = ck²

(i) (a + b + c) (a – b + c) = a² + b² + c²
L.H.S. = (a + b + c) (a – b + c)
= (ck² + ck + c) (ck² – ck+ c)
= c(k² + k + 1)c(k² – k + 1)
= c²(k⁴ + k² + 1)
R.H.S. = a² + b² + c² = c²k⁴ + c²k² + c²
= c²(k⁴ + k² + 1)
∴ L.H.S. = R.H.S.

(ii) \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)

(iii) \(\frac{a^3+b^3+c^3}{a^2 b^2 c^2}=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)

(iv) (4a² + lab + 9b²) : (4b² + 7bc + 9c²) = a : c

Question 4.
If a, b, c, d are in continued proportion, prove that :
(i) (b – c)² + (c – a)² + (d – b)² = (a – d)²
(ii) \(\sqrt{\frac{a^5+b^2 c^2+a^3 c^2}{b^4 c+d^4+b^2 c d^2}}=\frac{a}{d}\)
(iii) \(\sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)}\)
(iv) \(\frac{3 a+5 d}{5 a+7 d}=\frac{3 a^3+5 b^3}{5 a^3+7 b^3}\)
(v) \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
(vi) pa³ + qb³ + rc³ : pb³ + qc³ + rd³ = a : d
Solution:
∵ a, b, c, d are in continued proportion
∴ \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = \(\frac { c }{ d }\) = k (Suppose)
∴ c = dk, b = ck = dk.k = dk²
a = bk = dk²k = dk³
(i) (b – c)² + (c – a)² + (d – b)² = (a – d)²
L.H.S. = (b – c)² + (c – a)² + (d – b)²
= (dk² – dk)² + (dk – dk³)² + (d – dk²)²
= d²k² (k – 1)² + d²k²(1 – k²) + d²(1 – k²)²
= d²k² (1 -k)² + d²k² (1 – k)² (1 + k)² + d² (1 + k)²(1 – k)²
= d² (1 – k²) [k² + k² (1 + k)² + (1 + k)²]
= d² (1 – k²) [k² + k² (1 + 2k + k²) + 1 + 2k + k²]
= d² (1 – k²) [k² + k² + 2k³ + k⁴ + 1 + 2k + k²]
= d²(1 – k²)(k⁴ + 2k³ + 3k² + 2k + 1)
R.H.S. = (a – d)² = (dk³ – d)²
= d² (k³ – 1)² = d² (1 – k³)²
= d²(1 – k)²(1 + k + k²)²
= d²(1 – k)²[1 + k² + k⁴ + 2k + 2k³ + 2k²]
= d²(1 – k)² (k⁴ + 2k³ + 3k² + 2k + 1)
∴ L.H.S. = R.H.S.

(ii) \(\sqrt{\frac{a^5+b^2 c^2+a^3 c^2}{b^4 c+d^4+b^2 c d^2}}=\frac{a}{d}\)

(iii) \(\sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)}\)

(iv) \(\frac{3 a+5 d}{5 a+7 d}=\frac{3 a^3+5 b^3}{5 a^3+7 b^3}\)

(v) \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
L.H.S = \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
= \(\frac{d^3 k^9+d^3 k^6+d^3 k^3}{d^3 k^6+d^3 k^3+d^3}\)
= \(\frac{d^3 k^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)}\) = k³
R.H.S = \(\frac { a }{ d }\) = \(\frac { dk³ }{ d }\) = k³
∴ L.H.S. = R.H.S.

(vi) pa³ + qb³ + rc³ : pb³ + qc³ + rd³ = a : d

Question 5.
If \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\), prove that \(\frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c)(y-z)}\) + \(\frac{c z-a x}{(c+a)(z-x)}\) = 3
Solution:

Question 6.
If \(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\), show that each ratio is equal to \(\frac{x+y+z}{a+b+c}\).
Solution:
\(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\)
= \(\frac{x+y+z}{b+c-a+c+a-b+a+b-c}\)
= \(\frac{x+y+z}{a+b+c}\)
Hence Proved.

Question 7.
If \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\), prove that a (b – c) + b (c – a) + c (a – b) = 0.
Solution:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\) (suppose)
∴ a = k (b + c), b = k (c + a), c = k (a + b)
Now,
L.H.S. = a (b – c) + b (c – a) + c (a – b)
= k (b + c) (b – c) + k (c + a) (c – a) + k (a + b)(a – b)
= k (b² – c²) + k(c² – a²) + k (a² – b²)
= k (b² – c² + c² – a² + a² – b²) = k x 0 = 0
= R.H.S.
Hence proved.

Question 8.
If ax = by = cz, prove that \(\frac{x^2}{y z}+\frac{y^2}{z x}+\frac{z^2}{x y}=\frac{b c}{a^2}+\frac{c a}{b^2}+\frac{a b}{c^2}\).
Solution:

Self Evaluation and Revision (Latest ICSE Questions)

Question 1.
Given \(\frac { a }{ b }\) = \(\frac { c }{ d }\), prove that \(\frac{3 a-5 b}{3 a+5 b}\) = \(\frac{3 c-5 d}{3 c+5 d}\).
Solution:
\(\frac { a }{ b }\) = \(\frac { c }{ d }\) = k (suppose)
∴ a = bk, c = dk
L.H.S = \(\frac{3 a-5 b}{3 a+5 b}=\frac{3 b k-5 b}{3 b k+5 b}=\frac{b(3 k-5)}{b(3 k+5)}\)
= \(\frac{3 k-5}{3 k+5}\)
R.H.S = \(\frac{3 c-5 d}{3 c+5 d}=\frac{3 d k-5 d}{3 d k+5 d}\)
= \(\frac{d(3 k-5)}{d(3 k+5)}=\frac{3 k-5}{3 k+5}\)
∴ L.H.S = R.H.S

Question 2.
Two numbers are in the ratio of 3 : 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.
Solution:
Ratio between two numbers = 3 : 5
Let first number be = 3x
Then second number = 5x
Adding 8 to each, we get
\(\frac{3 x+8}{5 x+8}\) = \(\frac { 2 }{ 3 }\) ⇒ 2 (5x + 8) = 3 (3x + 8)
⇒ 10x + 16 = 9x + 24
⇒ 10x – 9x = 24 – 16
⇒ x = 8
∴ First number = 3x = 3 x 8 = 24
and second number = 5x = 5 x 8 = 40

Question 3.
If a : b = 5 : 3, find (5a + 8b) : (6a – 7b).
Solution:
a : b = 5 : 3
or \(\frac { a }{ b }\) = \(\frac { 5 }{ 3 }\)
Now (5a + 8b) : (6a – 7b)
= \(\frac{5 a+8 b}{6 a-7 b}=\frac{5 \frac{a}{b}+8 \frac{b}{b}}{6 \frac{a}{b}-7 \frac{b}{b}}\) (Dividing by b)
= \(\frac{5 \frac{a}{b}+8}{6 \frac{a}{b}-7}=\frac{5 \frac{5}{3}+8}{6 \frac{5}{3}-7}\)
= \(\frac{\frac{25}{3}+8}{\frac{30}{3}-7}=\frac{\frac{25+24}{3}}{\frac{30-21}{3}}=\frac{\frac{49}{3}}{\frac{9}{3}}\)
= \(\frac{49}{3} \times \frac{3}{9}=\frac{49}{9}\)
∴ (5a + 8b) : (6a – 7b) = 49 : 9

Question 4.
If \(\frac{3 a+4 b}{3 c+4 d}=\frac{3 a-4 b}{3 c-4 d}\) prove that \(\frac { a }{ b }\) = \(\frac { c }{ d }\).
Solution:
\(\frac{3 a+4 b}{3 c+4 d}=\frac{3 a-4 b}{3 c-4 d}\)
⇒ \(\frac{3 a+4 b}{3 a-4 b}=\frac{3 c+4 d}{3 c-4 d}\) (By altemendo)
Applying componendo and dividendo,
\(\frac{3 a+4 b+3 a-4 b}{3 a+4 b-3 a+4 b}=\frac{3 c+4 d+3 c-4 d}{3 c+4 d-3 c+4 d}\)
⇒ \(\frac{6 a}{8 b}=\frac{6 c}{8 d} \Rightarrow \frac{a}{b}=\frac{c}{d}\) (Dividing by \(\frac { 6 }{ 8 }\))
Hence proved.

Question 5.
The work done by (x – 3) men in (2x + 1) days and the work done by (2x + 1) men in (x + 4) days are in the ratio 3 : 10. Find the value of x.
Solution:
Work done by (x – 3) men in (2x + 1) days = (x – 3) (2x + 1)
and work done by (2x + 1) men in (x + 4) days = (2x + 1) (x + 4)
Now according to the given condition
\(\frac{(x-3)(2 x+1)}{(2 x+1)(x+4)}\) = \(\frac { 3 }{ 10 }\)
⇒ \(\frac { x-3 }{ x+4 }\) = \(\frac { 3 }{ 10 }\)
⇒ 10x – 30 = 3x + 12
⇒ 10x – 3x = 12 + 30 ⇒ 7x = 42
∴ x = 6

Question 6.
What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion?
Soluition:
Let x be subtracted from each number, we get
23 – x, 30 – x, 57 – x and 78 – x
According to the condition that these will be proportional, then
\(\frac{23-x}{30-x}=\frac{57-x}{78-x}\)
⇒ (23 – x) (78 – x) = (57 – x) (30 – x)
⇒ 1794 – 23x- 78x + x² = 1710 – 57x – 30x + x²
⇒ 1794 – 101x + x² = 1710 – 87x + x²
⇒ x² – 101x – x² + 87x = 1710 – 1794
⇒ – 14x = – 84
⇒ x = \(\frac { -84 }{ -14 }\) = 6
Required number = 6

Question 7.
What number must be added to each of the numbers 6, 15, 20 and 43 to make them in proportion.
Solution:
Let x be added to each of the numbers 6,15, 20 and 43, then
6 + x, 15 + x, 20 + x and 43 + x will be in proportion
∴ \(\frac{6+x}{15+x}=\frac{20+x}{43+x}\)
⇒ (6 + x) (43 + x) = (20 + x) (15 + x)
⇒ 258 + 6x + 43x + x² = 300 + 20x + 15x + x²
⇒ 258 + 49x + x² = 300 + 35x + x²
⇒ 49x + x² – 35x – x² = 300 – 258
⇒ 14x = 42
⇒ x = \(\frac { 42 }{ 14 }\) = 3
∴ Required number = 3

Question 8.
If \(\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}\), find x : y.
Solution:
\(\frac{3 x+5 y}{3 x-5 y}=\frac{7}{3}\)
Applying componendo and dividendo
\(\frac{3 x+5 y+3 x-5 y}{3 x+5 y-3 x+5 y}=\frac{7+3}{7-3}\)
\(\frac{6 x}{10 y}=\frac{10}{4} \Rightarrow \frac{x}{y}=\frac{10}{4} \times \frac{10}{6}\)
⇒ \(\frac{x}{y}=\frac{25}{6}\)
∴ x : y = 25 : 6

Question 9.
If x = \(\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}\), prove that 3bx² – 2ax + 3b = 0
Solution:

Question 10.
If \(\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d}\), prove that \(\frac { a }{ b }\) = \(\frac { c }{ d}\).
Solution:
\(\frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d} \Rightarrow \frac{8 a+5 b}{8 a-5 b}=\frac{8 c+5 d}{8 c-5 d}\)
Now applying componendo and dividendo
\(\frac{8 a+5 b+8 a-5 b}{8 a+5 b-8 a+5 b}=\frac{8 c+5 d+8 c-5 d}{8 c+5 d-8 c+5 d}\)
\(\frac{16 a}{10 b}=\frac{16 c}{10 d} \Rightarrow \frac{a}{b}=\frac{c}{d}\) (Dividing by \(\frac { 16 }{ 10 }\))
Hence \(\frac { a }{ b }\) = \(\frac { c }{ d }\)

Question 11.
What least number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion?
Solution:
Let x be added to 5, 11, 19 and 37 to make them in proportion.
∴ 5 + x : 11 + x : : 19 + x : 37 + x
⇒ (5 + x) (37 + x) = (11 + x) (19 + x)
⇒ 185 + 5x + 37x + x² = 209 + 11x + 19x + x²
⇒ 185 + 42x + x² = 209 + 30x + x²
⇒ 42x – 30x + x² – x² = 209 – 185
⇒ 12x = 24
⇒ x = 2
∴ Least number to be added = 2

Question 12.
Given \(\frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62}\) that using componendo and dividendo find a : b.
Solution:
Given : \(\frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62}\)
By componendo and dividendo
\(\frac{a^3+3 a b^2+b^3+3 a^2 b}{a^3+3 a b^2-b^3-3 a^2 b}=\frac{63+62}{63-62}\)
⇒ \(\frac{(a+b)^3}{(a-b)^3}=\left(\frac{5}{1}\right)^3\)
⇒ \(\frac{a+b}{a-b}\) = 5
⇒ a + b = 5a – 5b
⇒ 4a – 6b = 0
⇒ \(\frac { a }{ 2 }\) = \(\frac { 3 }{ 2 }\)
∴ a : b = 3 : 2

Question 13.
If x, y, z are in continued proportion, prove that \(\frac{(x+y)^2}{(y+x)^2}=\frac{x}{z}\).
Solution:
If x, y, z are in continued proportion ⇒ y² = xz
L.H.S. = \(\frac{(x+y)^2}{(y+z)^2}=\frac{x^2+y^2+2 x y}{y^2+z^2+2 y z}\)
= \(\frac{x^2+x z+2 x y}{x z+z^2+2 y z}\)
= \(\frac{x(x+z+2 y)}{z(x+z+2 y)}=\frac{x}{z}\)

Question 14.
Given x = \(\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}\) use componendo and dividendo to prove that b² = \(\frac{2 a^2 x}{x^2+1}\)
Solution:
If \(\frac { x }{ 1 }\) = \(\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}\)
Applying componendo and dividendo both sides

Question 15.
Using componendo and dividendo, find the value of x
\(\frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+4}-\sqrt{3 x-5}}\) = 9
Solution:
\(\frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+4}-\sqrt{3 x-5}}\) = 9
By componendo and dividendo, we have

Question 16.
6 is the mean proportion between two numbers x and y and 48 is the third proportional of x andy. Find the numbers.
Solution:
Mean propertion = \(\sqrt{xy}\)
⇒ 6 = \(\sqrt{xy}\)
⇒ xy = 36 (squaring both sides)
⇒ y = \(\frac { 36 }{ x }\) … (i)
For 3rd proportion :
x : y :: y : 48
⇒ \(\frac{x}{y}=\frac{y}{48}\) ⇒ y² = 48x
⇒ (\(\frac { 36 }{ x }\))² = 48x [using (i)]
⇒ 48x³ = 36 x 36 ⇒ x³ = \(\frac{36 \times 36}{48}\)
⇒ x³ = 27 = (3)³ ⇒ x = 3
From equation (i) :
y = \(\frac { 36 }{ x }\) = \(\frac { 36 }{ 3 }\) = 12
Hence, x = 3 and y = 12

Question 17.
If x = \(\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}\), using properties of proportion, show that x² – 2ox + 1 = 0.
Solution:
x = \(\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}\)
Applying componendo and dividendo
\(\frac { 1 }{ 2 }\)
We have x = \(\frac{x+1}{x-1}=\frac{\sqrt{a+1}+\sqrt{a-1}+\sqrt{a+1}-\sqrt{a-1}}{\sqrt{a+1}+\sqrt{a-1}+\sqrt{a+1}+\sqrt{a-1}}\)
⇒ \(\frac{x+1}{x-1}=\frac{2 \sqrt{a+1}}{2 \sqrt{a-1}}\)
(Applying componendo and dividendo)
Now, by taking square on both sides,

Question 18.
Using the properties of proportion, solve for x, given \(\frac{x^4+1}{2 x^2}=\frac{17}{8}\)
Solution:
\(\frac{x^4+1}{2 x^2}=\frac{17}{8}\)
\(\frac{x^4+1+2 x^2}{x^4+1-2 x^2}=\frac{17+8}{17-8}\)
(Applying componendo and dividendo)
= \(\frac{\left(x^2+1\right)^2}{\left(x^2-1\right)^2}=\frac{25}{9}\)
Taking square root on both sides
= \(\frac{x^2+1}{x^2-1}=\frac{5}{3}\)
= \(\frac{x^2+1+x^2-1}{x^2+1-\left(x^2-1\right)}=\frac{5+3}{5-3}\)
(By componendo and dividendo)
= \(\frac { 2x² }{ 2 }\) = \(\frac { 8 }{ 2 }\)
x² = 4
x = ± 2

Question 19.
If \(\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}\), then find the value of:
(i) x : y
(ii) \(\frac{x^3+y^3}{x^3-y^3}\)
Solution:

Question 20.
If a, b, c are in continued proportion, prove that
(a + b + c) (a – b + c) = a² + b² + c².
Solution:
Given that a, b and c are in continued proportion
∴ \(\frac { a }{ b }\) = \(\frac { b }{ c }\)
∴ b² = ac
To prove : (a + b + c) (a – b + c) = a² + b² + c²
L.H.S. = (a + b + c) (a – b + c)
= a(a – b + c) + b(a – b + c) + c(a – b + c)
= a² – ab + ac + ab – b² + bc + ac – bc + c²
= a² + ac – b² + ac + c²= a² + b² – b² + b² + c² [∵ b² = ac]
= a² + b² + c²
= R.H.S.

Question 21.
Give \(\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}\). Find x : y.
Solution:
Given that \(\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}\)
Using componendo-dividendo, we have

Thus the required ratio is x : y = 2 : 3

Question 22.
If \(\frac { x }{ a }\) = \(\frac { y }{ b }\) = \(\frac { z }{ c }\) show that \(\frac{x^3}{a^3}+\frac{y^3}{b^3}\) + \(\frac{z^3}{c^3}=\frac{3 x y z}{a b c}\).
Solution: