Question

Question 6
Write the following cubes in expanded form:

$\left(i\right)(2x+1{)}^3(ii\left)\right(2a-3b{)}^3\left(iii\right){[\frac{3}{2}x+1]}^3\left(iv\right){[x-\frac{2}{3}y]}^3$

Answer 1

$\left(i\right)(2x+1{)}^{3}Usingidentity(a+b{)}^3=a^3+b^3+3ab(a+b)=(2x{)}^3+1^3+3(2x\left)\right(1\left)\right(2x+1)=8x^3+1+6x(2x+1)=8x^3+12x^2+6x+1$

$\left(ii\right)(2a-3b{)}^{3}Usingidentity(a-b{)}^3=a^3-b^3-3ab(a-b)=(2a{)}^3-(3b{)}^3-3\times 2a\times 3b(2a-3b)=8a^3-27b^3-18ab(2a-3b)=8a^3-27b^3-36a^{2}b+54a{b}^2$

$\left(iii\right){[\frac{3}{2}x+1]}^{3}Usingidentity(a+b{)}^3=a^3+b^3+3ab(a+b)={\left(\frac{3}{2}x\right)}^3+1^3+3\left(\frac{3}{2}x\right)(1)(\frac{3}{2}x+1)=\frac{27}{8}{x}^3+1+\frac{9}{2}x(\frac{3}{2}x+1)=\frac{27}{8}{x}^3+\frac{27}{4}{x}^2+\frac{9}{2}x+1$

$\left(iv\right){[x-\frac{2}{3}y]}^{3}Usingidentity(a-b{)}^3=a^3-b^3-3ab(a-b)=x^3-{\left(\frac{3}{2}y\right)}^3-3(x)\left(\frac{2}{3}y\right)(x-\frac{2}{3}y)=x^3-\frac{8}{27}{y}^3-2xy(x-\frac{2}{3}y)=x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}x{y}^2$