Алгебра: Sin^4a+cos^4a-sin^6a-cos^6a=sin^2a cos^2a докажите тождество

Решите задачу:

$sin^4(a)+cos^4(a)-sin^6(a)-cos^6(a)=sin^2(a)cos^2(a)\\(1-sin^2(a))sin^4(a)+cos^4(a)-cos^6(a)=sin^2(a)cos^2(a)\\cos^2(a)+sin^2(a)=1--\ \textgreater \ 1-sin^2(a)=cos^2(a)\\cos^2(a)sin^4(a)+cos^4(a)-cos^6(a)=sin^2(a)cos^2(a)\\(sin^4(a)+cos^2(a)-cos^4(a))cos^2(a)=sin^2(a)cos^2(a)\\a^2-b^2=(a-b)(a+b)\\((sin^2(a)-cos^2(a))(sin^2(a)+cos^2(a))+cos^2(a))cos^2(a)=\\=sin^2(a)cos^2(a)\\(-(cos^2(a)-sin^2(a))+cos^2(a))cos^2(a)=sin^2(a)cos^2(a)\\(-cos(2a)+cos^2(a))cos^2(a)=sin^2(a)cos^2(a)\\cos(2a)=cos^2(a)-sin^2(a)$
$(-(cos^2(a)-sin^2(a))+cos^2(a))cos^2(a)=sin^2(a)cos^2(a)\\\\(-cos^2(a)+sin^2(a)+cos^2(a))cos^2(a)=sin^2(a)cos^2(a)\\sin^2(a)cos^2(a)=sin^2(a)cos^2(a)$