To prove the trigonometric equation:
[tex]\sin ^{3} x+\cos ^{3} x=(\sin x+\cos x)(1-\sin x \cos x)[/tex]
[tex]RHS=(\sin x+\cos x)(1-\sin x \cos x)[/tex]
We know that [tex]\sin^2x +\cos^2x=1[/tex], substitute this in place of 1.
[tex]=(\sin x+\cos x)(\sin^2x +\cos^2x-\sin x \cos x)[/tex]
Multiply each term of the first term with each term of the 2nd term.
[tex]=\sin^3x + \sin x \cos^2x-\sin^2 x \cos x+\cos x \sin^2 x + \cos^3 x-\sin x\cos^2 x[/tex]
Group like terms together.
[tex]=\sin^3x +( \sin x \cos^2x-\sin x\cos^2 x)+(\cos x \sin^2 x-\sin^2 x \cos x) + \cos^3 x[/tex]
[tex]=\sin^3x +( 0)+(0) + \cos^3 x[/tex]
[tex]=\sin^3x + \cos^3 x[/tex]
= LHS
RHS = LHS
[tex]\sin ^{3} x+\cos ^{3} x=(\sin x+\cos x)(1-\sin x \cos x)[/tex]
Hence proved.
