Question

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Answer 1

The give polynomial is

$P(x) = 3x^{2}+2x-6$

Since $\alpha$ and $\beta$ are the zeroes of P(x), by the relation between zeroes and coefficients, we get

$\alpha+\beta=-\frac{2}{3}$

$\alpha\beta=\frac{-6}{3}=-2$

(i)

Now, $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta$

$\implies (\alpha-\beta)^{2}=(-\frac{2}{3})^{2}-4(-2)$

$\implies (\alpha-\beta)^{2}=\frac{4}{9}+8=\frac{76}{9}$

$\implies \boxed{\alpha-\beta=\pm \frac{2\sqrt{19}}{3}}$

(ii)

Now, ${\alpha}^{2}+{\beta}^{2}=(\alpha+\beta)^{2}-2\alpha\beta$

$\implies {\alpha}^{2}+{\beta}^{2}=(-\frac{2}{3})^{2}-2(-2)$

$\implies {\alpha}^{2}+{\beta}^{2}=\frac{4}{9}+4$

$\implies \boxed{{\alpha}^{2}+{\beta}^{2}=\frac{40}{9}}$

(iii)

Now, ${\alpha}^{3}+{\beta}^{3}=(\alpha+\beta)({\alpha}^{2}+{\beta}^{2}-\alpha\beta)$

$\implies {\alpha}^3+{\alpha}^{3}=-\frac{2}{3}(\frac{40}{9}+2)$

$\implies {\alpha}^{3}+{\beta}^{3}=-\frac{2}{3}\times \frac{58}{9}$

$\implies \boxed{{\alpha}^{3}+{\beta}^{3}=-\frac{116}{27}}$

(iv)

Now, $\frac{1}{\alpha}+\frac{1}{\beta}$

$=\frac{\beta+\alpha}{\alpha\beta}$

$\implies \frac{1}{\alpha}+\frac{1}{\beta}=\frac{-\frac{2}{3}}{-2}$

$\implies \boxed{\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{3}}$

(v)

Now, ${\alpha}^{3}{\beta}^{3}=(\alpha\beta)^{3}$

$\implies {\alpha}^{3}{\beta}^{3}=(-2)^{3}$

$\implies \boxed{{\alpha}^{3}{\beta}^{3}=-8}$

Answer 2

$p(x) = 3 {x}^{2} + 2x - 6$

$\alpha \:and\: \beta$ are zeroes of the polynomial.

Therefore :-

$sum \: of \: zeroes= \frac{ - b}{a} = \frac{ - 2}{3} \\ \\ \\product \: of \: zeroes = \frac{c}{a} = \frac{ - 6}{3} \\ = ( - 2)$

$\rule{200}{2}$

Now,

$\boxed{ \red{ \sf{(i) \alpha - \beta}}}$

${( \alpha - \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta \\ \\ \\ {( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta \\ \\ \\ {( \alpha - \beta )}^{2} = {( \frac{ - 2}{3})}^{2} - 4( - 2) \\ \\ \\ {( \alpha - \beta )}^{2} = \frac{4}{9} + 8 = \frac{4 + 72}{9} \\ \\ {( \alpha - \beta )}^{2} = \frac{76}{9} \\ \\ \\ \alpha - \beta = \pm\sqrt{ \frac{76}{9} } \\ \\ \\ \alpha - \beta = \pm \frac{2}{3} \sqrt{19}$

$\rule{200}{2}$

$\boxed{ \orange{(ii) { \alpha }^{2} + { \beta }^{2}}}$

${ \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ \\ \\ { \alpha }^{2} + { \beta }^{2} = {( \frac{ - 2}{3})}^{2} - 2( - 2) \\ \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{4}{9} + 4 \\ \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{4 + 36}{9} \\ \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{40}{9}$

$\rule{200}{2}$

$\boxed{ \green{(iii) { \alpha }^{3} + { \beta }^{3}}}$

${ \alpha }^{3} + { \beta }^{3} =( \alpha + \beta )( { \alpha }^{2} + { \beta }^{2} - \alpha \beta ) \\ \\ \\ = (\dfrac{-2}{3})(\dfrac{40}{9} - (-2))\\ \\ \\=\dfrac{-2}{3} \times \dfrac{58}{9}=\dfrac{-116}{27}$

$\rule{200}{2}$

$\boxed{\red{(iv) \dfrac{1}{\alpha}+\dfrac{1}{\beta}}}$

$\frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta } \\ \\ \\ = \frac{ \frac{ - 2}{3} }{ - 2} = \frac{1}{3}$

$\rule{200}{2}$

$\boxed{\blue{\alpha^{3} \beta^{3}}}$

${ \alpha }^{3}{ \beta }^{3} = {( \alpha \beta )}^{3} \\ \\ \\ = {( - 2)}^{3} = ( - 8)$

$\rule{200}{2}$