Solve frac{a_{1}(1-q^3)}{1-q}=a_{1}*q+10a_{1}

$$\frac { a _ { 1 } ( 1 - q ^ { 3 } ) } { 1 - q } = a _ { 1 } \cdot q + 10 a _ { 1 }$$

Solve for a_1

$$\left\{\begin{matrix}a_{1}=0\text{, }&q\neq 1\\a_{1}\in \mathrm{R}\text{, }&|q|=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by -q+1.

$$a_{1}\left(1-q^{3}\right)=a_{1}q\left(-q+1\right)+10a_{1}\left(-q+1\right)$$

Use the distributive property to multiply a_{1} by 1-q^{3}.

$$a_{1}-a_{1}q^{3}=a_{1}q\left(-q+1\right)+10a_{1}\left(-q+1\right)$$

Use the distributive property to multiply a_{1}q by -q+1.

$$a_{1}-a_{1}q^{3}=-a_{1}q^{2}+a_{1}q+10a_{1}\left(-q+1\right)$$

Use the distributive property to multiply 10a_{1} by -q+1.

$$a_{1}-a_{1}q^{3}=-a_{1}q^{2}+a_{1}q-10a_{1}q+10a_{1}$$

Combine a_{1}q and -10a_{1}q to get -9a_{1}q.

$$a_{1}-a_{1}q^{3}=-a_{1}q^{2}-9a_{1}q+10a_{1}$$

Add a_{1}q^{2} to both sides.

$$a_{1}-a_{1}q^{3}+a_{1}q^{2}=-9a_{1}q+10a_{1}$$

Add 9a_{1}q to both sides.

$$a_{1}-a_{1}q^{3}+a_{1}q^{2}+9a_{1}q=10a_{1}$$

Subtract 10a_{1} from both sides.

$$a_{1}-a_{1}q^{3}+a_{1}q^{2}+9a_{1}q-10a_{1}=0$$

Combine a_{1} and -10a_{1} to get -9a_{1}.

$$-9a_{1}-a_{1}q^{3}+a_{1}q^{2}+9a_{1}q=0$$

Combine all terms containing a_{1}.

$$\left(-9-q^{3}+q^{2}+9q\right)a_{1}=0$$

The equation is in standard form.

$$\left(-q^{3}+q^{2}+9q-9\right)a_{1}=0$$

Divide 0 by -9-q^{3}+q^{2}+9q.

$$a_{1}=0$$

Solve for q

$$\left\{\begin{matrix}\\q=-3\text{; }q=3\text{, }&\text{unconditionally}\\q\neq 1\text{, }&a_{1}=0\end{matrix}\right.$$