Solve for: (4x^4-8y^3)^2

Expression: ${\left( 4{x}^{4}-8{y}^{3} \right)}^{2}$

Write as a product

$\left( 4{x}^{4}-8{y}^{3} \right) \times \left( 4{x}^{4}-8{y}^{3} \right)$

Multiply each term in the first parentheses by each term in the second parentheses (FOIL)

$4{x}^{4} \times 4{x}^{4}-4{x}^{4} \times 8{y}^{3}-8{y}^{3} \times 4{x}^{4}-8{y}^{3} \times \left( -8{y}^{3} \right)$

Calculate the product

$16{x}^{8}-4{x}^{4} \times 8{y}^{3}-8{y}^{3} \times 4{x}^{4}-8{y}^{3} \times \left( -8{y}^{3} \right)$

Calculate the product

$16{x}^{8}-32{x}^{4}{y}^{3}-8{y}^{3} \times 4{x}^{4}-8{y}^{3} \times \left( -8{y}^{3} \right)$

Calculate the product

$16{x}^{8}-32{x}^{4}{y}^{3}-32{x}^{4}{y}^{3}-8{y}^{3} \times \left( -8{y}^{3} \right)$

Multiplying two negatives equals a positive: $\left( - \right) \times \left( - \right)=\left( + \right)$

$16{x}^{8}-32{x}^{4}{y}^{3}-32{x}^{4}{y}^{3}+8{y}^{3} \times 8{y}^{3}$

Calculate the product

$16{x}^{8}-32{x}^{4}{y}^{3}-32{x}^{4}{y}^{3}+64{y}^{6}$

Collect like terms

$16{x}^{8}-64{x}^{4}{y}^{3}+64{y}^{6}$