Exercise
$\left(\frac{2}{3}x^2-y^3\right)^4$
Step-by-step Solution
Intermediate steps
1
Expand the binomial $\left(\frac{2}{3}x^2-y^3\right)^4$
$\left(\frac{2}{3}x^2\right)^4-4\left(\frac{2}{3}x^2\right)^3y^3+6\left(\frac{2}{3}x^2\right)^2y^{6}-\frac{8}{3}x^2y^{9}+y^{12}$
Intermediate steps
2
The power of a product is equal to the product of it's factors raised to the same power
$\frac{16}{81}x^{8}-4\cdot \left(\frac{8}{27}\right)x^{6}y^3+6\cdot \left(\frac{4}{9}\right)x^{4}y^{6}-\frac{8}{3}x^2y^{9}+y^{12}$
Intermediate steps
3
Multiply the fraction and term in $-4\cdot \left(\frac{8}{27}\right)x^{6}y^3$
$\frac{16}{81}x^{8}-\frac{32}{27}x^{6}y^3+6\cdot \left(\frac{4}{9}\right)x^{4}y^{6}-\frac{8}{3}x^2y^{9}+y^{12}$
Intermediate steps
4
Multiply the fraction and term in $6\cdot \left(\frac{4}{9}\right)x^{4}y^{6}$
$\frac{16}{81}x^{8}-\frac{32}{27}x^{6}y^3+\frac{8}{3}x^{4}y^{6}-\frac{8}{3}x^2y^{9}+y^{12}$
Final answer to the exercise
$\frac{16}{81}x^{8}-\frac{32}{27}x^{6}y^3+\frac{8}{3}x^{4}y^{6}-\frac{8}{3}x^2y^{9}+y^{12}$