Math, asked by maaltikumari, 9 months ago

if alpha and beta are the roots of the equation X²- 6 X + 6 = 0 what is alpha cube + beta cube + Alpha square + beta square + alpha + beta is equal to

Answers

Answered by MaheswariS
0

\underline{\textbf{Given:}}

$\boldsymbol{\alpha\;\&\beta}$ \textbf{are roots of}\;\mathsf{x^2-6x+6=0}

\underline{\textbf{To find:}}

\textsf{The value of}

$\boldsymbol{{\alpha}^3+{\beta}^3+{\alpha}^2+{\beta}^2+{\alpha}+{\beta}}$

\underline{\textbf{Solution:}}

\mathsf{Since}\;$\boldsymbol{\alpha}$\;\mathsf{and}\;$\boldsymbol{\beta}$\;\mathsf{are\;roots\;of\;x^2-6x+6=0}

\textsf{we have}

$\boldsymbol{\alpha+\beta=\dfrac{-(-6)}{1}\;\;and\;\;\alpha\,\beta=\dfrac{6}{1}}$

$\boldsymbol{\alpha+\beta=6\;\;and\;\;\alpha\,\beta=6}$

\mathsf{Now,}

\mathsf{$\boldsymbol{{\alpha}^3+{\beta}^3+{\alpha}^2+{\beta}^2+{\alpha}+{\beta}}$}

\textsf{Using the following identities,}

\boxed{\begin{minipage}{11cm}$\\\\\mathsf{(a+b)^3=a^3+b^3+3ab(a+b)\,\implies\,a^3+b^3=(a+b)^3-3ab(a+b)}\\\\\mathsf{(a+b)^2=a^2+b^2+2\,ab\,\implies\,a^2+b^2=(a+b)^2-2\,ab}\\$\end{minipage}}

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

\mathsf{=(6)^3-3(6)(6)+(6)^2-2\,(6)+6}

\mathsf{=216-108+36-12+6}

\mathsf{=108+30}

\mathsf{=138}

\implies\boxed{$\boldsymbol{{\alpha}^3+{\beta}^3+{\alpha}^2+{\beta}^2+{\alpha}+{\beta}=138}$}

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