Step 1: Understand what is being asked.
We expand $\cos 6x$ as a polynomial in $\cos x$ and read off the coefficients $a,b,c,d$ of $\cos^6x,\cos^4x,\cos^2x$ and the constant.
Step 2: Use a clever substitution instead of memorising.
Put $t=\cos^2x$. Then the right side is $at^3+bt^2+ct+d$, a cubic in $t$. We can find $a+b+c$ by smart choices of $t$.
Step 3: Pick $x=0$ to get the total.
At $x=0$, $\cos x=1$ so $t=1$, and $\cos 6x=\cos 0=1$. This gives $a+b+c+d=1$.
Step 4: Pick $x=\frac{\pi}{2}$ to isolate $d$.
At $x=\frac{\pi}{2}$, $\cos x=0$ so every $\cos$ term dies and only $d$ survives. Also $\cos 6x=\cos 3\pi=-1$, so $d=-1$.
Step 5: Subtract to get $a+b+c$.
From $a+b+c+d=1$ and $d=-1$, we get $a+b+c=1-(-1)=2$.
Step 6: Cross-check with the known identity.
The standard expansion $\cos 6x=32\cos^6x-48\cos^4x+18\cos^2x-1$ gives $a+b+c=32-48+18=2$, confirming our shortcut.
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