Question 154

If $$\frac{p^2}{q^2}+\frac{q^2}{p^2}$$=1 then the value of $$(p^{6}+q^{6})$$ is

Solution

Expression : $$\frac{p^2}{q^2}+\frac{q^2}{p^2}$$ = 1

=> $$\frac{p^{4}+q^{4}}{p^2q^2}$$ = 1

=> $$p^4+q^4 = p^2q^2$$ --------------Eqn(1)

Now, to find : $$(p^{6}+q^{6})$$

=> $$(p^2)^3 + (q^2)^3$$

Using the formula, $$a^3 + b^3 = (a+b)(a^2+b^2-ab)$$

=> $$(p^2+q^2)(p^4+q^4-p^2q^2)$$

From eqn (1), we get :

=> $$(p^2+q^2)(p^2q^2-p^2q^2)$$

=> $$(p^2+q^2)*0$$

= 0

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