Question 142

If $$a^2+a+1=0$$, then the value of $$a^4+a^2+1$$ is:

Solution

Given : $$a^2+a+1=0$$

Dividing both sides by $$'a'$$, we get

=> $$a+1+\frac{1}{a}=0$$

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

Squaring both sides, we get :

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

=> $$a^2+\frac{1}{a^2}+2(a)(\frac{1}{a})=1$$

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

=> $$a^4+a^2+1=0$$

=> Ans - (A)

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