Question

Find the cube of :

(i) $a+2$

(ii) $2a-1$

(iii) $2a+3b$

(iv) $3b-2a$

(v) $2x+\frac{1}{x}$

(vi) $x-\frac{1}{2}$

Answer 1

(i) $(a+2{)}^3=(a{)}^3+(2{)}^3+3\times a\times 2(a+2)=a^3+8+6a(a+2)=a^3+8+6a^2+12a=a^3+6a^2+12a+8$.

(ii) $(2a-1{)}^3=(2a{)}^3-(1{)}^3-3\times 2a\times 1(2a-1)=8a^3-1-6a(2a-1)=8a^3-1-12a^2+6a=8a^3-12a^2+6a-1$.

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

(iv) $(3b-2a{)}^3=(3b{)}^2-(2a{)}^3-3\times 3b\times 2a(3b-2a)=27b^3-8a^3-18ab(3b-2a)=27b^3-8a^3-54a{b}^2+36a^{2}b=27b^3-54b^{2}a+36b{a}^2-8a^3$.

(v) ${(2x+\frac{1}{x})}^3=(2x{)}^3+{\left(\frac{1}{x}\right)}^3+3\times 2x\times \frac{1}{x}(2x+\frac{1}{x})=8x^3+\frac{1}{x^3}+6(2x+\frac{1}{x})=8x^3+\frac{1}{x^3}+12x+\frac{6}{x}=8x^3+12x+\frac{6}{x}+\frac{1}{x^3}$.

(vi) ${(x-\frac{1}{2})}^3=(x{)}^3-{\left(\frac{1}{2}\right)}^2-3\times x\times \frac{1}{2}(x-\frac{1}{2})=x^3-\frac{1}{8}-\frac{3x}{2}(x-\frac{1}{2})=x^3-\frac{1}{8}-\frac{3x^2}{2}+\frac{3x}{4}=x^3-\frac{3x^2}{2}+\frac{3x}{4}-\frac{1}{8}$.