Question

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Answer 1

$\Large{\underline{\underline{\mathfrak{\green{\bf{Solution}}}}}}$

$\Large{\underline{\sf{\bf{Given}}}}$

$\mapsto\sf{\:\left(a+\dfrac{1}{a}\right)^2\:=\:3............(1)}$

$\Large{\underline{\sf{\bf{Find}}}}$

$\mapsto\sf{\:\left(a^3+\dfrac{1}{a^3}\right)}$

$\Large{\underline{\underline{\mathfrak{\green{\bf{Explanation}}}}}}$

Given,

$\mapsto\sf{\:\left(a+\dfrac{1}{a}\right)^2\:=\:3}$

$\:\:\:\:\:\small\sf{\:( x+y)^2=x^2+y^2+2xy}$

$\mapsto\sf{\:\left(a^2+\dfrac{1}{a^2}+2\times a\times \dfrac{1}{a}\right)\:=\:3}$

$\mapsto\sf{\:\left(a^2+\dfrac{1}{a^2}+2\right)\:=\:3}$

$\mapsto\sf{\:\left(a^2+\dfrac{1}{a^2}\right)\:=\:3\:-\:2}$

$\mapsto\sf{\:\left(a^2+\dfrac{1}{a^2}\right)\:=\:1.........(2)}$

_____________________

Again,

we have,

$\:\:\:\:\:\small\sf{\:( x^3+y^3)=(x+y)(x^2+y^2-xy)}$

$\mapsto\sf{\:\left(a^3+\dfrac{1}{a^3}\right)}$

$\sf{\:=(a+\dfrac{1}{a})(a^2+\dfrac{1}{a^2}-a\times \dfrac{1}{a})}$

$\sf{\:=(a+\dfrac{1}{a})(a^2+\dfrac{1}{a^2}-1)}$

$\:\:\:\:\:\small\sf{\:\left( Keep \:value\:by\:(1)\:and\:(2)\right)}$

$\sf{\:=(\sqrt{3})(1-1)}$

$\sf{\:=0\times (\sqrt{3})}$

$\sf{\bf{\:=0\:\:\:\:\:Ans}}$

______________________

Answer 2

Given:

$( a+\frac{1}{a})^{2} =3$

To prove:

$a^{3} +\frac{1}{a^3} =0$

Solution:

$\\\\a+\frac{1}{a}= \sqrt{3} \: \: ------------(i) \\\\\: \: ( a+\frac{1}{a})^{3} =(\sqrt{3 } )^3 \: \: [cubing\: both\: side ]\\\\a^3 + \frac{1}{a^3} +3.a.\frac{1}{a} (a+\frac{1}{a} )= 3\sqrt{3} \\\\a^3 + \frac{1}{a^3} +3(\sqrt{3} )=3\sqrt{3} \:\: [ from \: (i) ] \\\\a^3 + \frac{1}{a^3} = 3\sqrt{3}-3\sqrt{3}\\\\a^3 + \frac{1}{a^3} = 0$

Hence proved