Solution: 1) If $a=b$, we have $sin^6a+cos^6a+3sin^2acos^2a=1$, and this is true, strating from $(sin^2a+cos^2a)^3=1$ 2) If we have true the equality, we can write ${{\sin }^{6}}a+3{{\sin }^{2}}a{{\cos }^{6}}b=1\Leftrightarrow {{\sin }^{6}}a+3{{\sin }^{2}}a{{\cos }^{6}}b={{({{\sin }^{2}}a+{{\cos }^{2}}a)}^{3}}$, and from here we get $3{{\sin }^{2}}a({{\cos }^{2}}b-{{\cos }^{2}}a)+{{\cos }^{6}}a-{{\cos }^{6}}b=0$ and from here result that $cosa=cosb$, so $a=b$

gobathegreat wrote: Let $a$ and $b$ be real numbers from interval $\left[0,\frac{\pi}{2}\right]$. Prove that $$\sin^6 {a}+3\sin^2 {a}\cos^2 {b}+\cos^6 {b}=1$$if and only if $a=b$ Solution. 1). First recall formulas $(x+y)^3=x^3+y^3+3xy(x+y)$ and $\sin^2\alpha+\cos^2\alpha=1$. If $a=b$, then $$\sin^6 {a}+3\sin^2 {a}\cos^2a+\cos^6a=(\sin^2a+\cos^2a)^3-3\sin^2 {a}\cos^2a\left(\sin^2a+\cos^2a\right)+3\sin^2 {a}\cos^2a=1.$$2). Using formulas $(x+y)^3=x^3+y^3+3xy(x+y)$ and $x^3-y^3=(x-y)(x^2+xy+y^2)$, we get \begin{align*}0=&\sin^6 {a}+3\sin^2 {a}\cos^2 {b}+\cos^6 {b}-1\\ =&(\sin^2a+\cos^2b)^3-3\sin^2 {a}\cos^2 {b}\left(\sin^2a+\cos^2b\right)+3\sin^2 {a}\cos^2 {b}-1\\ =&(\sin^2a+\cos^2b)^3-1^3-3\sin^2 {a}\cos^2 {b}\left(\sin^2a+\cos^2b-1\right)\\ =&(\sin^2a+\cos^2b-1)\left[(\sin^2a+\cos^2b)^2+(\sin^2a+\cos^2b)+1-3\sin^2 {a}\cos^2 {b}\right]\\ =&(\sin^2a+\cos^2b-1)\left[\sin^4a+\cos^4b+\sin^2a+\cos^2b+\left(1-\sin^2 {a}\cos^2 {b}\right)\right], \end{align*}which gives $\sin^2a+\cos^2b-1=0$, leading to $\sin^2a=\sin^2b$ thereby $a=b$ since $a,b\in\left[0,\frac{\pi}{2}\right]$. $\blacksquare$

If a=b sin⁶a+3sin²a cos²a+cos⁶a (sin²a+cos²a)³=1 Proved