Originally Posted by

**Soroban** Hello, Annie!

Let: .$\displaystyle u \,= \,3x + 1$

Then we have: .$\displaystyle (u-6)u^2(u+6) + 68 \:=\:0\quad\Rightarrow\quad u^4-36u^2 + 68 \:=\:0$

. . which factors: .$\displaystyle (u^2-34)(u^2-2) \:=\:0$

. . and has roots: .$\displaystyle u \:=\:\pm\sqrt{34},\:\pm\sqrt{2}$

Back-substitute: .$\displaystyle \begin{Bmatrix}3x+1\:=\:\pm\sqrt{34} & \Rightarrow & x \:=\:\boxed{\frac{-1\pm\sqrt{34}}{3}} \\ \\

3x+1 \:=\:\pm\sqrt{2} & \Rightarrow & x \:=\:\boxed{\frac{-1\pm\sqrt{2}}{3}} \end{Bmatrix}$

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We can use a similar trick on #2 . . .

Let: .$\displaystyle u \:=\:x^2+6x+7$

Then we have: .$\displaystyle (u-1)(u+1) \:=\:528\quad\Rightarrow\quad u^2-529\:=\:0$

. . which factors: .$\displaystyle (u - 23)(u + 23) \:=\:0$

. . and has roots: .$\displaystyle u \:=\:\pm23$

Back-substitute:

$\displaystyle x^2+6x+7\:=\:23\quad\Rightarrow\quad x^2+6x-16\:=\:0$

. . $\displaystyle (x + 8)(x - 2)\:=\:0 \quad\Rightarrow\quad x \:=\:\boxed{-8,\:2}$

$\displaystyle x^2+6x+7\:=\:-23\quad\Rightarrow\quad x^2 + 6x + 30 \:=\:0$

. . $\displaystyle x \:=\:\frac{-6\pm\sqrt{-84}}{2} \;=\;\boxed{-3\pm i\sqrt{21}}$