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
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
