Question

if a and b are roots of equation x^2+x-7=0 then find a^3+b^3​

Answer 1

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Answer 2

SOLUTION

GIVEN

$\sf{ a \: and \: b \: are\:the \: roots \: of \: the \: equation }$

$\sf{ {x}^{2} + x - 7 = 0 \: }$

TO DETERMINE

$\sf{ {a}^{3} + {b}^{3} \: }$

CONCEPT TO BE IMPLEMENTED

$\sf{If \: \alpha \: and \: \beta \: are\:the \: roots \: of \: the \: equation }$

$\sf{a {x}^{2} + bx + c = 0 \: }$

Then

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

$\displaystyle \sf{ Product \: of \: the Zeros \: = \frac{c}{a} }$

EVALUATION

$\sf{ \because \: a \: and \: b \: are\:the \: roots \: of \: the \: equation }$

$\sf{ {x}^{2} + x - 7 = 0 \: }$

So

$\sf{a + b = - 1}$

$\sf{ab = - 7}$

Now we are aware of the identity that

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

$\implies \sf{ {a}^{3} + {b}^{3} = {( - 1)}^{3} - 3 \times ( - 1)( - 7) }$

$\implies \sf{ {a}^{3} + {b}^{3} = - 1 - 21}$

$\implies \sf{ {a}^{3} + {b}^{3} = - 22}$

FINAL ANSWER

$\boxed{ \: \sf{ {a}^{3} + {b}^{3} = - 22} \: \: }$

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