Question 126

If $$a + \left(\frac{1}{a}\right) = -1$$, then the value of $$(1 - a + a^2)(1 + a - a^2)$$ is

Solution

$$\ \frac{\ a^2+1}{a}=-1$$

$$a+\ \frac{\ 1}{a}=-1$$

$$\ \frac{\ a^2+1}{a}=-1$$

$$a^2=-a-1$$

Therefore

$$\left(1-a+a^2\right)\left(1+a-a^2\right)=\left(1-a-1-a\right)\left(1+a+1+a^2\right)$$=$$\left(-2a\right)\left(2+2a\right)$$=$$-4a-4a^2$$= $$-4\left(a+a^2\right)$$= -4(-1) = 4


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