The roots $$\alpha, \beta$$ of the equation $$3x^2 + \lambda x - 1 = 0$$, satisfy $$\cfrac{1}{\alpha^2} + \cfr

Question

The roots $$\alpha, \beta$$ of the equation $$3x^2 + \lambda x - 1 = 0$$, satisfy $$\cfrac{1}{\alpha^2} + \cfrac{1}{\beta^2} = 15$$. The value of $$(\alpha^3 + \beta^3)^2$$, is

Accepted Answer

4. From the sum and product of roots, we get: α + β = − λ 3 \alpha\ +\beta\ =-\dfrac{\lambda}{3} α + β = − 3 λ ​ and α β = − 1 3 \alpha\ \beta\ =-\dfrac{1}{3} α β = − 3 1 ​ Simplifying the expression given in the question, we get: α 2 + β 2 α 2 β 2 = 15 \dfrac{\alpha^2+\beta^2\ }{\alpha^2\beta^2\ }=15

The roots $\alpha, \beta$ of the equation $3x^2 + \lambda x - 1 = 0$ , satisfy $\cfrac{1}{\alpha^2} + \cfrac{1}{\beta^2} = 15$ .
The value of $(\alpha^3 + \beta^3)^2$ , is

The roots $\alpha, \beta$ of the equation $3x^2 + \lambda x - 1 = 0$ , satisfy $\cfrac{1}{\alpha^2} + \cfrac{1}{\beta^2} = 15$ .
The value of $(\alpha^3 + \beta^3)^2$ , is

The roots α,β\alpha, \betaα,β of the equation 3x2+λx−1=03x^2 + \lambda x - 1 = 03x2+λx−1=0, satisfy 1α2+1β2=15\cfrac{1}{\alpha^2} + \cfrac{1}{\beta^2} = 15α21​+β21​=15.The value of (α3+β3)2(\alpha^3 + \beta^3)^2(α3+β3)2, is

The roots α,β\alpha, \betaα,β of the equation 3x2+λx−1=03x^2 + \lambda x - 1 = 03x2+λx−1=0, satisfy 1α2+1β2=15\cfrac{1}{\alpha^2} + \cfrac{1}{\beta^2} = 15α21​+β21​=15.The value of (α3+β3)2(\alpha^3 + \beta^3)^2(α3+β3)2, is

The roots $\alpha, \beta$ of the equation $3x^2 + \lambda x - 1 = 0$ , satisfy $\cfrac{1}{\alpha^2} + \cfrac{1}{\beta^2} = 15$ .
The value of $(\alpha^3 + \beta^3)^2$ , is

The roots $\alpha, \beta$ of the equation $3x^2 + \lambda x - 1 = 0$ , satisfy $\cfrac{1}{\alpha^2} + \cfrac{1}{\beta^2} = 15$ .
The value of $(\alpha^3 + \beta^3)^2$ , is