If the sum of the squares of the reciprocals of the roots α and β of the equation 3x^2+λ x-1=0 is 15, then 6(α ^3+β ^3)^2 is equal to

Given $\frac{1}{{\alpha }^{2}}+\frac{1}{{\beta }^{2}}=15$α21​+β21​=15

For equation $3{x}^{2}+\lambda x-1=0$3x2+λx−1=0

$\alpha +\beta =\frac{-\lambda }{3}$α+β=3−λ​, $\alpha \beta =\frac{-1}{3}\Rightarrow {\alpha }^{2}{\beta }^{2}=\frac{1}{9}$αβ=3−1​⇒α2β2=91​

then ${\alpha }^{2}+{\beta }^{2}={(\alpha +\beta )}^{2}-2\alpha \beta$α2+β2=(α+β)2−2αβ$=\frac{{\lambda }^{2}}{9}-2(\frac{-1}{3})=\frac{{\lambda }^{2}}{9}+\frac{2}{3}$=9λ2​−2(3−1​)=9λ2​+32​

$$\Rightarrow {\alpha }^{2}+{\beta }^{2}=\frac{{\lambda }^{2}+6}{9}$$⇒α2+β2=9λ2+6​

Now $\frac{1}{{\alpha }^{2}}+\frac{1}{{\beta }^{2}}=15\Rightarrow \frac{{\alpha }^{2}+{\beta }^{2}}{{\alpha }^{2}{\beta }^{2}}=15$α21​+β21​=15⇒α2β2α2+β2​=15

$\Rightarrow \frac{{\lambda }^{2}+6}{9}\times \frac{1}{{\alpha }^{2}{\beta }^{2}}=15$⇒9λ2+6​×α2β21​=15 $\Rightarrow \frac{{\lambda }^{2}+6}{9}\times \frac{1}{\frac{1}{9}}=15$⇒9λ2+6​×91​1​=15

$\Rightarrow {\lambda }^{2}+6=15$⇒λ2+6=15 $\Rightarrow \lambda =\pm 3$⇒λ=±3

Now $6{({\alpha }^{3}+{\beta }^{3})}^{2}$6(α3+β3)2 $=6{((\alpha +\beta )({\alpha }^{2}-\alpha \beta +{\beta }^{2}))}^{2}$=6((α+β)(α2−αβ+β2))2

$$=6\times 1\times {(\frac{15}{9}+\frac{1}{3})}^{2}$$=6×1×(915​+31​)2

$$=6\times 1\times {(2)}^{2}=6\times 4=24$$=6×1×(2)2=6×4=24