Question

Expand using a formula.

(1) (2a + 3b + 3)²
(2) (x + 10) (x − 11)
(3) ${\left(3x+\frac{1}{3x}\right)}^3$
(4) (2m − n)³
(5) (2x + 3) (2x − 7)

Answer 1

(1) We know that ${\left(a+b+c\right)}^2=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)$
Thus, ${\left(2a+3b+3\right)}^2=4a^2+9b^2+9+12ab+18b+12a$

(2) (x + 10) (x − 11)
We know that $(x+a)(x+b)=x^2+(a+b)x+ab$
$$\begin{gathered} =x^2+(10-11)x+10\times (-11) \\ =x^2-x-110 \\ \therefore \left(x+10\right)\left(x-11\right)=x^2-x-110 \end{gathered}$$

(3) We know that ${\left(a+b\right)}^3=\left(a^3+b^3+3a^{2}b+3a{b}^2\right)$
Thus, ${\left(3x+\frac{1}{3x}\right)}^3=27x^3+\frac{1}{27x^3}+9x+\frac{1}{x}$

(4) We know that ${\left(a+b\right)}^3=\left(a^3+b^3+3a^{2}b+3a{b}^2\right)$
${\left(2m-n\right)}^3=8m^3-n^3-12m^{2}n+6m{n}^2$

(5) (2x + 3) (2x − 7)
We know that $(x+a)(x+b)=x^2+(a+b)x+ab$
$$\begin{gathered} =4x^2-\left(14-6\right)+3\times (-7) \\ =4x^2-8x-21 \\ \therefore \left(2x+3\right)\left(2x-7\right)=4x^2-8x-21 \end{gathered}$$