Question

Prove that:

$2(si{n}^{6}x+co{s}^{6}x)-3(si{n}^{4}x+co{s}^{4}x)+1=0$

Answer 1

LHS = $2(si{n}^{6}x+co{s}^{6}x)-3(si{n}^{4}x+co{s}^{4}x)+1$

$=2\left[\right(si{n}^{2}x{)}^3+\left(co{s}^{2}x{)}^3\right]-3(si{n}^{4}x+co{s}^{4}x)+1[\because a^3+b^3=(a+b\left)\right(a^2-ab+b^2\left)\right]=2\left[\right(si{n}^{2}x+co{s}^{2}x\left)\right(si{n}^{4}x-si{n}^{2}xco{s}^{2}x+co{s}^{4}x\left)\right]-3(si{n}^{4}x+co{s}^{4}x)+1=-[si{n}^{4}x+co{s}^{4}x+2si{n}^{2}xco{s}^{2}x]+1=-[si{n}^{2}x+co{s}^{2}x]+1$
=-1 + 1
=0
=RHS