Question

$2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1$ is equal to

• A. $2$
• B. $0$
• C. $4$
• D. $6$

Answer 1

The correct option is B $0$

$2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1$

Using $(a^3+b^3=(a+b{)}^3-3\cdot a\cdot b(a+b)\left)and\right(a^2+b^2=(a+b{)}^2-2ab)$, we get

$=2({({\sin }^2\theta +{\cos }^2\theta )}^3-3{\sin }^2\theta {\cos }^2\theta ({\sin }^2\theta +{\cos }^2\theta ))-3({({\sin }^2\theta +{\cos }^2\theta )}^2-2{\sin }^2\theta {\cos }^2\theta )+1$

$=2-6{\sin }^2\theta {\cos }^2\theta -3+6{\sin }^2\theta {\cos }^2\theta +1=0$