Q.Solve the equation $7x^3-43x^2=43x-7$

Rewrites the equation with all terms on one side, recognises it as a palindromic (reciprocal) cubic which always has $x=-1$ as a root, divides it out, and solves the resulting quadratic.

1. Rewrite $7x^3-43x^2=43x-7$ with all terms on one side: $7x^3-43x^2-43x+7=0$.
2. The coefficients, in order, are $7,\ -43,\ -43,\ 7$ — reading them forwards or backwards gives the same sequence. This is a palindromic (reciprocal-type) equation.
3. An odd-degree palindromic equation always has $x=-1$ as a root. Verify: $7(-1)^3-43(-1)^2-43(-1)+7=-7-43+43+7=0$ ✓.
4. Divide $7x^3-43x^2-43x+7$ by $(x+1)$ using synthetic/long division: $7x^3-43x^2-43x+7=(x+1)(7x^2-50x+7)$. (Check: $7x^2\cdot x=7x^3$; $7x^2\cdot1+(-50x)\cdot x=7x^2-50x^2=-43x^2$ ✓; $-50x\cdot1+7\cdot x=-50x+7x=-43x$ ✓; $7\cdot1=7$ ✓.) …