Question

if alpha,beta,gamma are the roots of equation x³+px²+qx+r=0 then sigma alpha²(beta+gama)​

Answer 1

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$\alpha , \beta\:and\:\gamma$ are the roots of equation ${x}^{3}+p{x}^{2}+qx+r=0$

Sum of roots taken one at a time=

$\sum\alpha=\alpha+\beta+\gamma=-p$

Sum of the roots taken two at a time =

$\sum\alpha\beta=\alpha\beta+\beta\gamma+\alpha\gamma=q$

Product of the roots= −r

$\sum{ \alpha }^{3} { \beta }^{3} = {( \alpha \beta )}^{3} + {( \beta \gamma )}^{3} + {( \alpha \gamma) }^{3}$

we know that,

${a}^{3} + {b}^{3} + {c}^{3} = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2}) - (ab + bc + ca) + 3abc$

${ (\alpha \beta) }^{3} + { (\beta \gamma )}^{3} + { (\alpha \gamma) }^{3} = ( \alpha \beta + \beta \gamma + \alpha \gamma ) \huge{(} \small { (\alpha \beta )}^{2} + {( \beta \gamma )}^{2} + {( \alpha \gamma )}^{2} - ( { \alpha \beta }^{2} \gamma + { \beta \gamma }^{2} \alpha + { \alpha \gamma }^{2} \beta ) \huge) \small + 3( \alpha \beta )( \beta \gamma )( \gamma \alpha ) \\ \\ =q \huge( \small {( \alpha \beta + \beta \gamma + \gamma \alpha) }^{2} - 2 { \alpha \beta }^{2} \gamma - 2 { \beta \gamma }^{2} \alpha - 2{ \gamma \alpha }^{2} \beta - { \alpha \beta }^{2} \gamma - { \beta \gamma }^{2} \alpha - { \gamma \alpha }^{2} \beta \huge) \small + 3{( \alpha \beta \gamma )}^{2} \\ \\ = (q)( {q} ^{2} - 3 \alpha \beta \gamma ( \alpha + \beta + \gamma )) + 3 {r}^{2} \\ \\ = (q)( {q}^{2} + 3rp) + 3 {r}^{2} \\ \\ = {q}^{3} + 3pqr + 3 {r}^{2}$

Answer 2

Step-by-step explanation:

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