Solve (a+bb+cc+a]=a

$$( a + b b + c c + a ] = a$$

Solve for a

$$a=-\left(b^{2}+c^{2}\right)$$

Solution Steps

Multiply b and b to get b^{2}.

$$a+b^{2}+cc+a=a$$

Multiply c and c to get c^{2}.

$$a+b^{2}+c^{2}+a=a$$

Combine a and a to get 2a.

$$2a+b^{2}+c^{2}=a$$

Subtract a from both sides.

$$2a+b^{2}+c^{2}-a=0$$

Combine 2a and -a to get a.

$$a+b^{2}+c^{2}=0$$

Subtract b^{2} from both sides. Anything subtracted from zero gives its negation.

$$a+c^{2}=-b^{2}$$

Subtract c^{2} from both sides.

$$a=-b^{2}-c^{2}$$

Solve for b (complex solution)

$$b=-i\sqrt{a+c^{2}}$$

$$b=i\sqrt{a+c^{2}}$$

Solve for b

$$b=\sqrt{-a-c^{2}}$$

$$b=-\sqrt{-a-c^{2}}\text{, }a\leq -c^{2}$$