Solve 8alphagamma^3+betadelta^3-8betagamma^3-alphadelta^3

$$8 \alpha \gamma ^ { 3 } + \beta \delta ^ { 3 } - 8 \beta \gamma ^ { 3 } - \alpha \delta ^ { 3 }$$

Factor

$$\left(\alpha -\beta \right)\left(2\gamma -\delta \right)\left(4\gamma ^{2}+2\gamma \delta +\delta ^{2}\right)$$

Solution Steps

Do the grouping 8\alpha \gamma ^{3}+\beta \delta ^{3}-8\beta \gamma ^{3}-\alpha \delta ^{3}=\left(8\alpha \gamma ^{3}-\alpha \delta ^{3}\right)+\left(-8\beta \gamma ^{3}+\beta \delta ^{3}\right), and factor out \alpha in the first and -\beta in the second group.

$$\alpha \left(8\gamma ^{3}-\delta ^{3}\right)-\beta \left(8\gamma ^{3}-\delta ^{3}\right)$$

Factor out common term 8\gamma ^{3}-\delta ^{3} by using distributive property.

$$\left(8\gamma ^{3}-\delta ^{3}\right)\left(\alpha -\beta \right)$$

Consider 8\gamma ^{3}-\delta ^{3}. Rewrite 8\gamma ^{3}-\delta ^{3} as \left(2\gamma \right)^{3}-\delta ^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

$$\left(2\gamma -\delta \right)\left(4\gamma ^{2}+2\gamma \delta +\delta ^{2}\right)$$

Rewrite the complete factored expression.

$$\left(\alpha -\beta \right)\left(2\gamma -\delta \right)\left(4\gamma ^{2}+2\gamma \delta +\delta ^{2}\right)$$

Evaluate

$$\left(\alpha -\beta \right)\left(8\gamma ^{3}-\delta ^{3}\right)$$