[Algebra]
Problem: Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$.
Determine all real numbers $k$ for which $f(x)$ is constant for all values of $x$.

Solution: Since $\sin ^{2} x+\cos ^{2} x=1$, then $\cos ^{2} x=1-\sin ^{2} x$, so

$$
\begin{aligned}
f(x) & =\sin ^{6} x+\left(1-\sin ^{2} x\right)^{3}+k\left(\sin ^{4} x+\left(1-\sin ^{2} x\right)^{2}\right) \\
& =\sin ^{6} x+1-3 \sin ^{2} x+3 \sin ^{4} x-\sin ^{6} x+k\left(\sin ^{4} x+1-2 \sin ^{2} x+\sin ^{4} x\right) \\
& =(1+k)-(3+2 k) \sin ^{2} x+(3+2 k) \sin ^{4} x
\end{aligned}
$$

Therefore, if $3+2 k=0$ or $k=-\frac{3}{2}$, then $f(x)=1+k=-\frac{1}{2}$ for all $x$ and so is constant. (If $k \neq-\frac{3}{2}$, then we get

$$
\begin{aligned}
f(0) & =1+k \\
f\left(\frac{1}{4} \pi\right) & =(1+k)-(3+2 k)\left(\frac{1}{2}\right)+(3+2 k)\left(\frac{1}{4}\right)=\frac{1}{4}+\frac{1}{2} k \\
f\left(\frac{1}{6} \pi\right) & =(1+k)-(3+2 k)\left(\frac{1}{4}\right)+(3+2 k)\left(\frac{1}{16}\right)=\frac{7}{16}+\frac{5}{8} k
\end{aligned}
$$

which cannot be all equal for any single value of $k$, so $f(x)$ is not constant if $k \neq-\frac{3}{2}$.)
Since $\sin ^{2} x+\cos ^{2} x=1$, then

$$
\begin{aligned}
f(x) & =\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)+k\left(\sin ^{4} x+\cos ^{4} x\right) \\
& =\left(\sin ^{4}+2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x-3 \sin ^{2} x \cos ^{2} x\right) \\
& \quad k\left(\sin ^{4} x+2 \sin ^{2} x \cos ^{2} x+\cos ^{4} x-2 \sin ^{2} x \cos ^{2} x\right) \\
& =\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x\right)+k\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right) \\
& =1-3 \sin ^{2} x \cos ^{2} x+k\left(1-2 \sin ^{2} x \cos ^{2} x\right) \\
& =(1+k)-(3+2 k) \sin ^{2} x \cos ^{2} x
\end{aligned}
$$

Therefore, if $3+2 k=0$ or $k=-\frac{3}{2}$, then $f(x)=1+k=-\frac{1}{2}$ for all $x$ and so is constant.
For $f(x)$ to be constant, we need $f^{\prime}(x)=0$ for all values of $x$.

Calculating using the Chain Rule,

$$
\begin{aligned}
f^{\prime}(x) & =6 \sin ^{5} x \cos x-6 \cos ^{5} x \sin x+k\left(4 \sin ^{3} x \cos x-4 \cos ^{3} x \sin x\right) \\
& =2 \sin x \cos x\left(3\left(\sin ^{4} x-\cos ^{4} x\right)+2 k\left(\sin ^{2} x-\cos ^{2} x\right)\right) \\
& =2 \sin x \cos x\left(\sin ^{2} x-\cos ^{2} x\right)\left(3\left(\sin ^{2} x+\cos ^{2} x\right)+2 k\right) \\
& =2 \sin x \cos x\left(\sin ^{2} x-\cos ^{2} x\right)(3+2 k)
\end{aligned}
$$

If $3+2 k=0$ or $k=-\frac{3}{2}$, then $f^{\prime}(x)=0$ for all $x$, so $f(x)$ is constant.

(If $3+2 k \neq 0$, then choosing $x=\frac{1}{6} \pi$ for example gives $f^{\prime}(x) \neq 0$ so $f(x)$ is not constant.)

Final answer: $-\frac{3}{2}$