Question

$x = \frac{a}{b} + \frac{b}{a}$
and
$y = \frac{a}{b} - \frac{b}{a}$
if show

$x {}^{4} + y {}^{4} - 2x {}^{2} y {}^{2} = 16$



​

Answer 1

Answer:

R=a+

)8(

-b

Step-by-step explanation:

Answer 2

Given :

• $\bf{x = \dfrac{a}{b} + \dfrac{b}{a}}$

• $\bf{y = \dfrac{a}{b} - \dfrac{b}{a}}$

To find :

To proof :-

$\bf{x^{4} + y^{4} - 2x^{2}y^{2} = 16}$

Solution :

$:\implies \bf{x^{4} + y^{4} - 2x^{2}y^{2} = 16}$

By using the identity , we get :

$\bf{(a - b)^{2} = a^{2} - 2ab - b^{2}}$

$:\implies \bf{(x^{2} + y^{2})^{2} = 16}$

By Substituting the value of x and y in the equation , we get :

$:\implies \bf{\bigg[\bigg(\dfrac{a}{b} + \dfrac{b}{a}\bigg)^{2} + \bigg(\dfrac{a}{b} - \dfrac{b}{a}\bigg)^{2}\bigg]^{2} = 16}$

Let $\bf{\dfrac{a}{b}}$ be $\bf{p}$ , so $\bf{\dfrac{b}{a}}$ will be $\bf{\dfrac{1}{p}}$

By Substituting them in the equation , we get :

$:\implies \bf{\bigg[\bigg(p + \dfrac{1}{p}\bigg)^{2} + \bigg(p - \dfrac{1}{p}\bigg)^{2}\bigg]^{2} = 16}$

By using the identity , we get :

• $\bf{(a + b)^{2} = a^{2} + 2ab - b^{2}}$

• $\bf{(a - b)^{2} = a^{2} - 2ab - b^{2}}$

$:\implies \bf{\bigg[\bigg(p^{2} + 2 \times p \times \dfrac{1}{p} + \dfrac{1}{p^{2}}\bigg) + \bigg(p^{2} - 2 \times p \times \dfrac{1}{p} + \dfrac{1}{p^{2}}\bigg)\bigg]^{2} = 16}$

$:\implies \bf{\bigg[\bigg(p^{2} + 2 \times \not{p} \times \dfrac{1}{\not{p}} + \dfrac{1}{p^{2}}\bigg) + \bigg(p^{2} - 2 \times \not{p} \times \dfrac{1}{\not{p}} + \dfrac{1}{p^{2}}\bigg)\bigg]^{2} = 16}$

$:\implies \bf{\bigg[\bigg(p^{2} + 2 + \dfrac{1}{p^{2}}\bigg) + \bigg(p^{2} - 2 + \dfrac{1}{p^{2}}\bigg)\bigg]^{2} = 16}$

$:\implies \bf{\bigg(p^{2} + 2 + \dfrac{1}{p^{2}} - p^{2} + 2 - \dfrac{1}{p^{2}}\bigg)^{2} = 16}$

$:\implies \bf{\bigg(\not{p^{2}} + 2 + \dfrac{1}{\not{p^{2}}} - \not{p^{2}} + 2 - \dfrac{1}{\not{p^{2}}}\bigg)^{2} = 16}$

$:\implies \bf{(2 + 2)^{2} = 16}$

$:\implies \bf{4^{2} = 16}$

$:\implies \bf{16 = 16}$

$\boxed{\therefore \bf{16 = 16}}$

Hence , LHS is equal to the RHS.

Thus it is proved that :

$\boxed{\bf{x^{4} + y^{4} - 2x^{2}y^{2} = 16}}$

Proved !