Question

If alpha and beta are the zeroes of the quadratic polynomial f(x)=6x²+x-2,find the value of alpha³+beta³

Answer 1

$\sf\red{\underline{\underline{Answer:-}}}$

$\sf{The \ value \ of \ \alpha^{3}+\beta^{3} \ is \ \frac{-37}{216}}$

$\sf\orange{Given:}$

$\sf{The \ given \ polynomial \ is}$

$\sf{\implies{f(x)=6x^{2}+x-2}}$

$\sf{\implies{Zeroes \ are \ \alpha \ and \ \beta}}$

$\sf\pink{To \ find:}$

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

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ polynomial \ is}$

$\sf{\implies{f(x)=6x^{2}+x-2}}$

$\sf{Here, \ a=6, \ b=1 \ and \ c=-2}$

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

$\sf{\therefore{\alpha+\beta=\frac{-1}{6}...(1)}}$

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

$\sf{\therefore{\alpha\beta=\frac{-2}{6}}}$

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

$\sf{According \ to \ identity}$

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

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

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

$\sf{\therefore{\alpha^{3}+\beta^{3}=(\frac{-1}{6})^{3}-3(\frac{-1}{3})(\frac{-1}{6})}}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=(\frac{-1}{6})^{3}-(\frac{1}{6})}}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{-1}{216}-\frac{1}{6}}}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{-1-36}{216}}}$

$\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{-37}{216}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ \alpha^{3}+\beta^{3} \ is \ \frac{-37}{216}}}}$

Answer 2

$\large\bf\underline \orange{Given:-}$

• p(x) = 6x² + x -2

$\large\bf\underline \orange{To \: find:-}$

• value of α³ + β³

$\huge\bf\underline \green{Solution:-}$

★ Given equation = 6x² + x -2

• a = 6
• b = 1
• c = -2

α and β are the roots of the given polynomial.

sum of zeroes = coefficient of x/coefficient of x²

• α + β = -b/a

⠀⠀⠀⠀⠀➝ α + β = -1/6

• α β = c/a

⠀⠀⠀⠀⠀➝ αβ = -2/6

⠀⠀⠀⠀⠀➝ α β = -1/3

$\pink{\boxed{ \bf({a}+ {b}) {}^{3} = {a}^{3} + b {}^{3} + 3ab (a + b)}} \\ \\ \longmapsto \rm \: ({a}+ {b}) {}^{3} = {a}^{3} + b {}^{3} + 3ab (a + b) \\ \\ \longmapsto \rm \: {a}^{3} + b {}^{3} = (a + b) {}^{3} - 3ab(a + b) \\ \\ \longmapsto \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = ( \alpha + \beta ) {}^{3} - 3 \alpha \beta ( \alpha + \beta ) \\ \\ \longmapsto \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =( \frac{ - 1}{6} ) {}^{3} - \cancel 3 \times \frac{ - 1}{ \cancel3} ( \frac{ - 1}{6} ) \\ \\ \longmapsto \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 1}{216} - \frac{1}{6} \\ \\ \longmapsto \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 1 - 36}{216} \\ \\ \longmapsto \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 37}{216}$

So,

α³ + β³ = -37/216