Sin^6x+cos^6x=1-3sin^2xcos^2x proof

I am ashamed,Can anyone help me prove the identity sin^6x+cos^6x=1-3sin^2xcos^2x ? ( plz)

a^6 + b^6 =>
(a^2)^3 + (b^2)^3 =>
(a^2 + b^2) * ((a^2)^2 - (a^2) * (b^2) + (b^2)^2) =>
(a^2 + b^2) * (a^4 - (ab)^2 + b^4)

(sin(x)^2 + cos(x)^2) * (sin(x)^4 - (sin(x)cos(x))^2 + cos(x)^4) =>
1 * (sin(x)^4 + cos(x)^4 - sin(x)^2 * cos(x)^2) =>
sin(x)^4 + cos(x)^4 - sin(x)^2 * cos(x)^2 =>
sin(x)^4 + (1 - sin(x)^2)^2 - sin(x)^2 * cos(x)^2 =>
sin(x)^4 + 1 - 2sin(x)^2 + sin(x)^4 - sin(x)^2 * cos(x)^2 =>
1 + 2sin(x)^4 - 2sin(x)^2 - sin(x)^2 * cos(x)^2 =>
1 + 2sin(x)^4 - sin(x)^2 * (2 + cos(x)^2) =>
1 + sin(x)^2 * (2sin(x)^2 - 2 - cos(x)^2) =>
1 + sin(x)^2 * (2 - 2cos(x)^2 - 2 - cos(x)^2) =>
1 + sin(x)^2 * (-3cos(x)^2) =>
1 - 3sin(x)^2 * cos(x)^2

tnx

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a^3+b^3 = (a+b)^3 - 3ab(a+b)
put a = sin ^2 t and b= cos^2 t

to get sin ^6 t + cos ^6 t= (sin ^2t+cos^2t)^3 - 3 sin ^2 t cos^2 t ( sin ^2 t + cos^2 t)
= 1 - 3 sin ^2 t cos^2 t

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tnx kali

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