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[tex]\Large{\underline{\underline{\mathfrak{\green{\bf{Solution}}}}}}[/tex]
[tex]\Large{\underline{\sf{\bf{Given}}}}[/tex]
[tex]\mapsto\sf{\:\left(a+\dfrac{1}{a}\right)^2\:=\:3............(1)}[/tex]
[tex]\Large{\underline{\sf{\bf{Find}}}}[/tex]
[tex]\mapsto\sf{\:\left(a^3+\dfrac{1}{a^3}\right)}[/tex]
[tex]\Large{\underline{\underline{\mathfrak{\green{\bf{Explanation}}}}}}[/tex]
Given,
[tex]\mapsto\sf{\:\left(a+\dfrac{1}{a}\right)^2\:=\:3}[/tex]
[tex]\:\:\:\:\:\small\sf{\:( x+y)^2=x^2+y^2+2xy}[/tex]
[tex]\mapsto\sf{\:\left(a^2+\dfrac{1}{a^2}+2\times a\times \dfrac{1}{a}\right)\:=\:3}[/tex]
[tex]\mapsto\sf{\:\left(a^2+\dfrac{1}{a^2}+2\right)\:=\:3}[/tex]
[tex]\mapsto\sf{\:\left(a^2+\dfrac{1}{a^2}\right)\:=\:3\:-\:2}[/tex]
[tex]\mapsto\sf{\:\left(a^2+\dfrac{1}{a^2}\right)\:=\:1.........(2)}[/tex]
_____________________
Again,
we have,
[tex]\:\:\:\:\:\small\sf{\:( x^3+y^3)=(x+y)(x^2+y^2-xy)}[/tex]
[tex]\mapsto\sf{\:\left(a^3+\dfrac{1}{a^3}\right)}[/tex]
[tex]\sf{\:=(a+\dfrac{1}{a})(a^2+\dfrac{1}{a^2}-a\times \dfrac{1}{a})}[/tex]
[tex]\sf{\:=(a+\dfrac{1}{a})(a^2+\dfrac{1}{a^2}-1)}[/tex]
[tex]\:\:\:\:\:\small\sf{\:\left( Keep \:value\:by\:(1)\:and\:(2)\right)}[/tex]
[tex]\sf{\:=(\sqrt{3})(1-1)}[/tex]
[tex]\sf{\:=0\times (\sqrt{3})}[/tex]
[tex]\sf{\bf{\:=0\:\:\:\:\:Ans}}[/tex]
______________________
Given:
[tex]( a+\frac{1}{a})^{2} =3[/tex]
To prove:
[tex]a^{3} +\frac{1}{a^3} =0[/tex]
Solution:
[tex]\\\\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\\[/tex]
Hence proved