Solve a/3+b^2-c^3+d^2=

$$\frac { a } { 3 } + b ^ { 2 } - c ^ { 3 } + d ^ { 2 } =$$

Evaluate

$$-c^{3}+b^{2}+d^{2}+\frac{a}{3}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply b^{2}-c^{3}+d^{2} times \frac{3}{3}.

$$\frac{a}{3}+\frac{3\left(b^{2}-c^{3}+d^{2}\right)}{3}$$

Since \frac{a}{3} and \frac{3\left(b^{2}-c^{3}+d^{2}\right)}{3} have the same denominator, add them by adding their numerators.

$$\frac{a+3\left(b^{2}-c^{3}+d^{2}\right)}{3}$$

Do the multiplications in a+3\left(b^{2}-c^{3}+d^{2}\right).

$$\frac{a+3b^{2}-3c^{3}+3d^{2}}{3}$$

Factor

$$\frac{a+3b^{2}-3c^{3}+3d^{2}}{3}$$

Solution Steps

Factor out \frac{1}{3}.

$$\frac{a+3b^{2}-3c^{3}+3d^{2}}{3}$$