Question

If p (x) = cx² + dx - c.
then find alpha³ + beta³​

Answer 1

HOPE THIS WILL HELP YOU

Answer 2

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

• p(x) = cx² + dx - c.

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

• value of α³ + β³

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

★ Given equation = cx² + dx - c

• a = c
• b = d
• c = -c

Let α and β are the roots of the given polynomial

• α + β = - b/a

⠀⠀⠀⠀⠀➝ - d/c

• α β = c/a

⠀⠀⠀⠀⠀➝ - c/c

⠀⠀⠀⠀⠀➝ -1

we know that,

$\pink{\boxed{ \bf({a}+ {b}) {}^{3} = {a}^{3} + b {}^{3} + 3ab (a + b)}} \\ \\ \longmapsto \rm \: ({a}+ {b}) {}^{3} = {a}^{3} + b {}^{3} + 3ab (a + b) \\ \\ \longmapsto \rm \: {a}^{3} + b {}^{3} = (a + b) {}^{3} - 3ab(a + b) \\ \\ \longmapsto \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = ( \alpha + \beta ) {}^{3} - 3 \alpha \beta ( \alpha + \beta ) \\ \\ \longmapsto \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (\frac{ - d }{c } ) {}^{3} - 3 \times - 1( \frac{ - d}{c} ) \\ \\ \longmapsto \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =( \frac{ - d}{c} ) {}^{3} - \frac{3d}{c} \\ \\ \longmapsto \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - d {}^{3} }{ {c}^{3} } - \frac{3d}{c} \\ \\ \longmapsto \bf\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - d {}^{3} - 3c {}^{2}d }{c {}^{3} }$

So,

Value of α³ + β³ =$\bf \frac{ - d {}^{3} - 3c {}^{2}d }{c {}^{3} }$