Write the following cubes in expanded form:
(i) (2x + 1)³ (ii) (2a – 3b) ³ (iii) [$\frac{3}{2}$ x + 1]³ (iv) [x - $\frac{2 }{ 3} $y]³

Question

Find the square root of $4a^2 + 9b^2 + c^2 + 12ab - 6bc - 4ac$.

Answer 1

- Cube of a sum: $ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $ - Cube of a difference: $ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $

Using $ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $ with $ a = 2x $, $ b = 1 $:

$ (2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + 1^3 $

$ 8x^3 + 12x^2 + 6x + 1 $

Using $ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $ with $ a = 2a $, $ b = 3b $:

$ (2a)^3 - 3(2a)^2(3b) + 3(2a)(3b)^2 - (3b)^3 $

$ 8a^3 - 36a^2b + 54ab^2 - 27b^3 $

Using $ (a+b)^3 $ formula with $ a = \frac{3}{2}x $, $ b = 1 $:

$ \left(\frac{3}{2}x\right)^3 + 3\left(\frac{3}{2}x\right)^2(1) + 3\left(\frac{3}{2}x\right)(1)^2 + 1^3 $

$ \frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1 $

Using $ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $ with $ a = x $, $ b = \frac{2}{3}y $:

$ x^3 - 3x^2\left(\frac{2}{3}y\right) + 3x\left(\frac{2}{3}y\right)^2 - \left(\frac{2}{3}y\right)^3 $

$ x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3 $

Write the following cubes in expanded form:
(i) (2x + 1)³ (ii) (2a – 3b) ³ (iii) [\(\frac{3}{2}\) x + 1]³ (iv) [x - \(\frac{2 }{ 3} \)y]³