Question

factorize: a^2/b^2+b^2/a^2​

Answer 1

Given :- factorize: a^2/b^2+b^2/a^2

Solution :- (for class 8 and 9)

→ (a²/b²) + (b²/a²)

→ (a/b)² + (b/a)²

using 1/a^m = a^(-m)

→ (a/b)² + (a/b)^(-2)

taking common ,

→ (a/b)²[1 + (a/b)^(-4)]

→ (a/b)²[1 + (b/a)⁴] (Ans.)

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Solution :- using iota (i) :-

→ (a²/b²) + (b²/a²)

using i² = (-1)

→ (a²/b²) - (i²)(b²/a²)

→ (a/b)² - (i)²(b/a)²

using (a² - b²) = (a + b)(a - b)

→ (a/b + ib/a)(a/b - ib/a) (Ans.)

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If we need to solve the a^2/b^2+b^2/a^2 :-

→ (a²/b²) + (b²/a²)

taking LCM ,

→ (a²*a² + b²*b²) /a²b²

using a^m * a^n = a^(m + n) in numerator ,

→ (a⁴ + b⁴)/a²b²

Let a² = x and b² = y

so,

→ x² + y² = (x + y)² - 2xy

→ a⁴ + b⁴ = (a² + b²)² - 2a²b²

then,

→ {(a² + b²)² - 2a²b²}/a²b²

→ [(a² + b²)²/a²b² - 2] (Ans.)

Learn more :-

Let a, b and c be non-zero real numbers satisfying (a³)/(b³ + c³) + (b³)/(c³ + a³) + (c³)/(a³ + b³)

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if a²+ab+b²=25

b²+bc+c²=49

c²+ca+a²=64

Then, find the value of

(a+b+c)² - 100 = __

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Answer 2

SOLUTION

TO FACTORISE

$\displaystyle \sf{ \frac{ {a}^{2} }{ {b}^{2} } + \frac{ {b}^{2} }{ {a}^{2} } }$

EVALUATION

Here the given expression is

$\displaystyle \sf{ \frac{ {a}^{2} }{ {b}^{2} } + \frac{ {b}^{2} }{ {a}^{2} } }$

We factorise it as below

$\displaystyle \sf{ \frac{ {a}^{2} }{ {b}^{2} } + \frac{ {b}^{2} }{ {a}^{2} } }$

$\displaystyle \sf{ = \frac{ {a}^{2} }{ {b}^{2} } - {i}^{2} \frac{ {b}^{2} }{ {a}^{2} } \: \: \: \: \: \: \: ( \because \: {i}^{2} = - 1) }$

$\displaystyle \sf{ = { \bigg( \frac{a}{b} \bigg)}^{2} - {i}^{2} { \bigg( \frac{b}{a} \bigg)}^{2} }$

$\displaystyle \sf{ = { \bigg( \frac{a}{b} \bigg)}^{2} - { \bigg( i \frac{b}{a} \bigg)}^{2} }$

$\displaystyle \sf{ = \bigg( \frac{a}{b} + i \frac{b}{a}\bigg) \bigg( \frac{a}{b} - i \frac{b}{a} \bigg) }$

FINAL ANSWER

$\boxed{ \: \: \: \displaystyle \sf{ \frac{ {a}^{2} }{ {b}^{2} } + \frac{ {b}^{2} }{ {a}^{2} }= \bigg( \frac{a}{b} + i \frac{b}{a}\bigg) \bigg( \frac{a}{b} - i \frac{b }{a} \bigg) } \: \: \: }$

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