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Laws of Exponents for Real Numbers

If c2+c=-1 ...

Question

If $c^{2} + c = - 1$ then the value of $c^{3} - 1$ is

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Solution

The correct option is B 0
Given $c^{2} + c = - 1$
$\Rightarrow c^{2} + c + 1 = 0$ Multiply by c
Then $c^{3} + c^{2} + c = 0$ add-1 both sides
$c^{3} + c^{2} + c - 1 = - 1$
$\Rightarrow c^{3} - 1 + c^{2} + c = - 1$
$\Rightarrow c^{3} - 1 + (- 1) = - 1$
$\Rightarrow c^{3} - 1 = 0$

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