Question

If a is not equal to 0 and a -1 by a =3 find (1) a square+1 by a square(2) a qube -1 by a qube

Answer 1

Taking a=X

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

$\tt{a\neq{0}}\\{X-\dfrac{1}{X}=3}$

${\sf{\green{\underline{\large{To\:find}}}}}$

$\tt{(1)X^2+\frac{1}{x^2}}\\(2){X^3-\dfrac{1}{X^3}}$

${\sf{\pink{\underline{\Large{Explanation}}}}}$

We know that

$\boxed{\boxed{\sf\red{(a-b)^2=a^2+b^2-2ab}}}$

Therefore

$\tt{(X-\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}$

Solving

$\tt{X-\dfrac{1}{X}=3}$

Both side squaring

$\tt{(X-\dfrac{1}{X})^2}$=(3)²

$\implies\tt{(X-\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}=9$

$\implies\tt{9=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}$

$\implies\tt{9=X^2+\dfrac{1}{X^2}-2}$

$\implies\tt{9+2=X^2+\dfrac{1}{X^2}}$

$\implies\tt{11=X^2+\dfrac{1}{X^2}}$

Now

$\tt{X^3+\dfrac{1}{X^3}}$

Formula used

$\implies\tt{X^3-\dfrac{1}{X^3)}=(X-\dfrac{1}{x})(X^2+\dfrac{1}{X^2}+\frac{1}{X}\times{X)}}$

¶utting value

$\implies\tt{X^3-\dfrac{1}{X^3}=3+11+1}$

$\implies\tt{X^3-\dfrac{1}{X^3}=15}$

$\implies\tt{X^3-\dfrac{1}{X^3}=15}$