Question

if (a+1/a)^2 =3 than show a^3 + 1/a^3=0​

Answer 1

Answer:

• Given that $\sf\bigg(a+\dfrac{1}{a}\bigg)^2=3$

• Show that $\sf a^3+\dfrac{1}{a^3}=0$

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf \sf\bigg(a+\dfrac{1}{a}\bigg)^2=3\\\\\\:\implies\sf a+\dfrac{1}{a} = \sqrt{3}\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Cubing both sides - }}\\\\:\implies\sf \bigg(a+\dfrac{1}{a}\bigg)^3 =( \sqrt{3})^3\\\\\\:\implies\sf {a}^{3} + \dfrac{1}{{a}^{3}} +3 \times a \times \dfrac{1}{a}\bigg(a+\dfrac{1}{a}\bigg) =3 \sqrt{3}\\\\\\:\implies\sf {a}^{3} + \dfrac{1}{{a}^{3}} +3\bigg(a+\dfrac{1}{a}\bigg) =3 \sqrt{3}\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{Putting \:value \:of \:\bigg(a+\dfrac{1}{a}\bigg)}} \\\\:\implies\sf{a}^{3} + \dfrac{1}{{a}^{3}} +3 \times \sqrt{3} =3 \sqrt{3}\\\\\\:\implies\sf {a}^{3} + \dfrac{1}{{a}^{3}} +3 \sqrt{3} =3 \sqrt{3}\\\\\\:\implies\sf {a}^{3} + \dfrac{1}{{a}^{3}} =3 \sqrt{3} -3 \sqrt{3}\\\\\\:\implies \underline{\boxed{\sf {a}^{3} + \dfrac{1}{{a}^{3}} = 0}}$

$\rule{200}{1}$

Shortcut for this :

1. Whenever the value of $\sf\bigg(a+\dfrac{1}{a}\bigg)^2=3$ then value of $\sf\bigg(a^3+\dfrac{1}{a^3}\bigg)=0$

⠀

2. Whenever the value of $\sf\bigg(a+\dfrac{1}{a}\bigg)^2=3$ then value of $\sf(a^6+1)=0$

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: Answer:-}}}}}$

◉ GIVEN :-

$: \longrightarrow{ \tt{ \bold{ {(a + \frac{1}{a}) }^{2} = 3}}}$

◉ TO SHOW:-

$: \longrightarrow{ \tt{ \bold{ \: a^3 + \frac{1}{a^3} =0}}}$

◉ FORMULA USED:-

$\longmapsto \: using \: (a + b)^3 \: = a^3 + b^3 + \: 3ab(a + b)</p><p>$

$\longmapsto \: a^3 + b^3 = (a + b)^3 - 3ab(a + b)$

Here we are using

a = a

b = 1/a

◉ SOLUTION:-

${\boxed{\underline{\red{L.H.S}}}}$

Now substituting the values.

$\leadsto \: a^3+(1/a)^3 \: = (a+ \frac{1}{a} )^3-3 \times a \times \frac{1}{a} (a+ \frac{1}{a} )$

$\leadsto \: \sqrt{3} ^3-3 \times \sqrt{3}$

$\leadsto \: 3 \sqrt{3} -3 \sqrt{3}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \longrightarrow{ \boxed{ \underline{ \mathtt{ \purple{0}}}}}$

${\boxed{\underline{\red{R.H.S}}}}$

HENCE PROVED :)