Question 113

If $$a^{2}+1=a$$, then the value of $$a^{12}+a^{6}+1$$ is :

Solution

Expression : $$a^{2}+1=a$$

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

Multiplying by $$(a+1)$$ on both sides

=> $$(a+1)(a^2-a+1) = 0$$

=> $$a^3 + 1^3 = 0$$

=> $$a^3 = -1$$

To find : $$a^{12}+a^{6}+1$$

= $$(a^3)^4 + (a^3)^2 + 1$$

= $$(-1)^4 + (-1)^2 + 1$$

= $$1+1+1 = 3$$

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