Question

Answer 1

a) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)

b) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^8-\dfrac{1}{27}\)

c) \(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)=\left(x^3-8\right)\left(x^3+8\right)=x^6-64\)

d) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-\left(8x^3-y^3\right)\\ =2y^3\)