Question

If alpha and beta are the zeroes of quadratic polynomial ax^2+bx+c, then evaluate:

(i)1/alpha^3 + 1/beta^3
(ii) (alpha^2/beta)+(beta^2/alpha)

Please write the answer...​

Answer 1

If α and β are Zeroes of a Quadratic Polynomial, ax² + bx + c then :

●  $\mathsf{Sum\;of\;the\;Zeroes\;(\alpha + \beta) = \dfrac{-b}{a}}$

●  $\mathsf{Product\;of\;the\;Zeroes\;(\alpha.\beta) = \dfrac{c}{a}}$

$\mathrm{(i).\;\;Given :\;\dfrac{1}{\alpha^3} + \dfrac{1}{\beta^3}}$

$\mathrm{\implies \dfrac{\alpha^3 + \beta^3}{\alpha^3.\beta^3}}$

Consider : (α + β)³

$:\implies$  (α + β)³ = α³ + 3α²β + 3αβ² + β³

$:\implies$  (α + β)³ = α³ + 3αβ(α + β) + β³

$:\implies$  α³ + β³ = (α + β)³ - 3αβ(α + β)

$\mathrm{\implies \dfrac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{(\alpha\beta)^3}}$

$\mathrm{\implies \dfrac{\bigg(\dfrac{-b}{a}\bigg)^3 - 3\bigg(\dfrac{c}{a}\bigg)\bigg(\dfrac{-b}{a}\bigg)}{\bigg(\dfrac{c}{a}\bigg)^3}}$

$\mathrm{\implies \dfrac{\dfrac{-b^3}{a^3} + \dfrac{3bc}{a^2}}{\dfrac{c^3}{a^3}}}$

$\mathrm{\implies \dfrac{\dfrac{3abc -b^3}{a^3}}{\dfrac{c^3}{a^3}}}$

$\mathrm{\implies \dfrac{{3abc -b^3}}{{c^3}}}$

$\mathrm{(ii).\;\;Given :\;\dfrac{\alpha^2}{\beta} + \dfrac{\beta^2}{\alpha}}$

$\mathrm{\implies \dfrac{\alpha^3 + \beta^3}{\alpha\beta}}$

$\mathrm{\implies \dfrac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta}}$

$\mathrm{\implies \dfrac{\bigg(\dfrac{-b}{a}\bigg)^3 - 3\bigg(\dfrac{c}{a}\bigg)\bigg(\dfrac{-b}{a}\bigg)}{\dfrac{c}{a}}}$

$\mathrm{\implies \dfrac{\dfrac{-b^3}{a^3} + \dfrac{3bc}{a^2}}{\dfrac{c}{a}}}$

$\mathrm{\implies \dfrac{\dfrac{3abc -b^3}{a^3}}{\dfrac{c}{a}}}$

$\mathrm{\implies \bigg[\dfrac{3abc -b^3}{a^3}\times\dfrac{a}{c}}\bigg]}$

$\mathrm{\implies \bigg[\dfrac{3abc -b^3}{a^2.c}\bigg]}$