\begin{align*} v^2 & = u^2 + 2as \\ -u^2 & = 2as - v^2 \\ u^2 & = v^2 - 2as \\ u & = \pm \sqrt{v^2 - 2as} \end{align*}

Note: Since $2^2 = 4$ and $(-2)^2 = 4$, the two possible values of the square root of $4$ are $2$ and $-2$. Thus, if $u^2 = v^2 - 2as$, then $u = \pm \sqrt{v^2 - 2as}$.

\begin{align*} y & = (7a + 6b)^3 - 8 \\ y + 8 & = (7a + 6b)^3 \\ \sqrt[3]{y + 8} & = 7a + 6b \\ -7a & = 6b - \sqrt[3]{y + 8} \\ 7a & = \sqrt[3]{y + 8} - 6b \\ a & = \frac{ \sqrt[3]{y + 8} - 6b }{7} \end{align*}

Note: We do not need the $\pm$ sign when taking the cube root. This is because $(3)^3 = 27$ but $(-3)^3 = -27$. Thus, if $(7a + 6b)^3 = y + 8 $, then $7a + 6b = \sqrt[3]{y + 8}$.

\begin{align*} y & = \sqrt{3x + 2z} + z \\ y - z & = \sqrt{3x + 2z} \\ (y - z)^2 & = 3x + 2z \qquad \qquad [\text{Square both sides}] \\ \underbrace{(y)^2 - 2(y)(z) + (z)^2}_{(a - b)^2 = a^2 - 2ab + b^2} & = 3x + 2z \\ y^2 - 2yz + z^2 & = 3x + 2z \\ -3x & = - y^2 - z^2 + 2z + 2yz \\ 3x & = y^2 + z^2 - 2z - 2yz \\ x & = \frac{y^2 + z^2 - 2z - 2yz}{3} \end{align*}

Note: Squaring is the opposite of square rooting. Thus, if $y - z = \sqrt{3x + 2z}$, then $(y - z)^2 = 3x + 2z$.

\begin{align*} \frac{y}{2} & = \sqrt[3]{4x - 5z} \\ \left(\frac{y}{2}\right)^3 & = 4x - 5z \qquad \qquad \quad [\text{Cube both sides}] \\ \\ \frac{y^3}{2^3} & = 4x - 5z \\ \frac{y^3}{8} & = \frac{4x - 5z}{1} \\ y^3 & = 8(4x - 5z) \qquad \qquad [\text{Cross-multiply}] \\ y^3 & = 32x - 40z \\ -32x & = -40z - y^3 \\ 32x & = 40z + y^3 \\ x & = \frac{40z + y^3}{32} \end{align*}

Note: Cubing is the opposite of cube rooting. Thus, if $\frac{y}{2} = \sqrt[3]{4x - 5z} \\$, then $ \left(\frac{y}{2}\right)^3 = 4x - 5z $.