Question

If α, β and γ are zeroes of the polynomial:-
p(x) = (x - 1)(x² + x + 3).

Then, find the value of
(α³ + β³ + γ³)
________...

Class IX - CBSE Question (Maths.).​

Answer 1

Answer:

$:\implies\sf p(x) =(x-1)(x^2+x+3)\\\\\\:\implies\sf p(x)=x(x^2+x+3)-1(x^2+x+3)\\\\\\:\implies\sf p(x) = x^3 + x^2 + 3x - x^2 - x - 3\\\\\\:\implies\sf p(x) = x^3 + 2x - 3 \\\\\bigstar\:\textsf{Here \: we \: have :} \\\\\textsf{ a = 1, \quad b = 0, \quad c = 2, \quad d = - \:3}$

⠀⠀⠀$\rule{160}{1}$

$\underline{\bigstar\:\textsf{Let's Consider a Formula :}}$

$\dashrightarrow\sf\:\:(\alpha)^3+(\beta)^3+(\gamma)^3 - 3\alpha \beta \gamma =( \alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta \gamma - \gamma \alpha)\\\\\\\dashrightarrow\sf\:\:(\alpha)^3+(\beta)^3+(\gamma)^3 - 3\alpha \beta \gamma =\dfrac{-\:b}{a} \times (\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta \gamma - \gamma \alpha)\\\\\\\dashrightarrow\sf\:\:(\alpha)^3+(\beta)^3+(\gamma)^3 - 3\alpha \beta \gamma =\dfrac{0}{1} \times (\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta \gamma - \gamma \alpha)\\\\\\\dashrightarrow\sf\:\:(\alpha)^3+(\beta)^3+(\gamma)^3 - 3\alpha \beta \gamma = 0\\\\\\\dashrightarrow\sf\:\:(\alpha)^3+(\beta)^3+(\gamma)^3 = 3\alpha \beta \gamma\\\\\\\dashrightarrow\sf\:\:(\alpha)^3+(\beta)^3+(\gamma)^3 = 3\times\dfrac{-\:d}{a}\\\\\\\dashrightarrow\sf\:\:(\alpha)^3+(\beta)^3+(\gamma)^3 = 3\times\dfrac{-\:( - 3)}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf (\alpha)^3+(\beta)^3+(\gamma)^3 = 9}}$

⠀

$\therefore\:\underline{\textsf{Hence, the required value will be \textbf{9}}}.$

$\rule{180}{2}$

$\boxed{\begin{minipage}{5.5 cm} { \bigstar\: \textsf{For a Cubic Polynomial :}}\\\\ {\qquad\sf p(x) = ax ^3 +\: bx^2 + cx + d}\\\sf with zeroes \alpha,\:\beta \:\sf and \:\gamma \\\\ {\textcircled{\footnotesize1}} \:\:\alpha +\beta + \gamma = \dfrac{ - \:b}{a} \\ \\{\textcircled{\footnotesize2}} \: \:\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} \\\\ {\textcircled{\footnotesize3}} \:\:\alpha \beta \gamma = \dfrac{ - \:d}{a} \end{minipage}}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\dagger \: \: \: {\sf{ p(x) = (x - 1)( {x}^{2} + x + 3) }}$

${\bf{\blue{\underline{To\:Find:}}}}$

$\dagger \: \: {\sf{ \alpha ^{3} + { \beta}^{3} + { \gamma}^{3} }}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies {\sf{ p(x) = (x - 1)( {x}^{2} + x + 3) }}$

$: \implies {\sf{ (x - 1)( {x}^{2} + x + 3) }}$

$: \implies {\sf{ {x}^{3} + {x}^{2} + 3x - {x}^{2} - x - 3}}$

$: \implies {\sf{ {x}^{3} + 0 {x}^{2} + 2x - 3}}$

Compare it with,

ax³+bx²+cx+d=0

___________________________________

$\dagger \: \boxed{\sf{sum \: of \: roots = \frac{coeff. \: of \: {x}^{n - 1} }{ coeff. \: {x}^{n} } = \frac{ - b}{a} }}$

$: \implies{\sf{ \alpha+ \beta +\gamma= - \frac{0}{1} }}$

$: \implies{\sf{ \alpha+ \beta +\gamma = 0 }}$

__________________________________

$\dagger \: \boxed{\sf{Pair \: Product = \frac{coeff. \: of \: {x}^{n - 2} }{ coeff. \: {x}^{n} } = \frac{ c}{a} }}$

$: \implies{\sf{ \alpha \beta + \beta \gamma + \gamma \beta= \frac{(2)}{1} }}$

$: \implies{\sf{ \alpha \beta + \beta \gamma + \gamma \beta= 2}}$

__________________________________

$\dagger \: \boxed{\sf{ Product \: of \: roots= ( - 1) ^{n} \frac{constant \: term }{ coeff. \: {x}^{n} } = \frac{ - d}{a} }}$

$: \implies{\sf{ \alpha \beta \gamma = - \frac{( - 3)}{1} }}$

$: \implies{\sf{ \alpha \beta \gamma = 3 }}$

__________________________________

Now According to the ques,

$\implies\: \: {\sf{ \alpha ^{3} + { \beta}^{3} + { \gamma}^{3} = 3 \alpha \beta \gamma \: \: \: \: \: \: ( \because \: \alpha + \beta + \gamma = 0)}}$

$\implies\: \: {\sf{ \alpha ^{3} + { \beta}^{3} + { \gamma}^{3} = 3 (3)}}$

$\implies\: \: {\sf{ \alpha ^{3} + { \beta}^{3} + { \gamma}^{3} = 9}}$