Question

3x²-2x+4=0 find 1/alpha cube+1/beta cube​

Answer 1

Answer:

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Answer 2

Answer:-

$\sf{The \ value \ of \ \frac{1}{\alpha^{3}}+\frac{1}{\beta^{3}} \ is \ -1.}$

Given:

• The given quadratic equation is
• $\sf{3x^{2}-2x+4=0}$

To find:

The value of $\sf{\frac{1}{\alpha^{3}}+\frac{1}{\beta^{3}}}$

Solution:

The given quadratic equation is

$\sf{3x^{2}-2x+4=0}$

$\sf{Here \ a=3, \ b=-2 \ and \ c=4}$

$\sf{Sum \ of \ roots=\frac{-b}{a}}$

$\sf{\therefore{\alpha+\beta=\frac{2}{3}...(1)}}$

$\sf{Product \ of \ roots=\frac{c}{a}}$

$\sf{\therefore{\alpha\beta=\frac{4}{3}...(2)}}$

$\sf{According \ to \ the \ identity.}$

$\boxed{\sf{a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)}}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3(\alpha\beta)(\alpha+\beta)}}$

$\sf{...from \ (1) \ and \ (2)}$

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

$\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{8}{27}-\frac{8}{3}}}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{8-72}{27}}}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{-64}{27}...(3)}}$

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$\sf{\frac{1}{\alpha^{3}}+\frac{1}{\beta^{3}}}$

$\sf{\implies{\frac{\alpha^{3}+\beta^{3}}{(\alpha\beta)^{3}}}}$

$\sf{...from \ (2) \ and \ (3)}$

$\sf{\implies{\frac{\frac{-64}{27}}{(\frac{4}{3})^{3}}}}$

$\sf{\implies{\frac{\frac{-64}{27}}{\frac{64}{27}}}}$

$\sf{\implies{\frac{-64}{27}\times\frac{27}{64}}}$

$\sf{\implies{-1}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ \frac{1}{\alpha^{3}}+\frac{1}{\beta^{3}} \ is \ -1.}}}$