Question

Write the following cubes in expanded form:
(A) ${\left(2x+1\right)}^3$
(B) $(2a-3b{)}^3$
(C) ${\left[{\displaystyle \frac{3}{2}}x+1\right]}^3$
(D) ${\left[x-{\displaystyle \frac{2}{3}}y\right]}^3$

Answer 1

Hint: Use the expansion formula of ${\left(a+b\right)}^3$ . Then substitute the given values in the formula to solve the above question. Be careful while substituting the values and look after the positive and negative signs while using the expansion formula.

Complete step-by-step answer:
(A) ${\left(2x+1\right)}^3$
We know that,
 ${\left(x+y\right)}^3=x^3+y^3+3xy(x+y)$
Here, $(2x+1{)}^3$
By comparing the expression with the formula we have, we can write
 $\Rightarrow x=2x,y=1$
Now, by substituting the values in the given formula, we get
 $(2x+1{)}^3={\left(2x\right)}^3+{\left(1\right)}^3+\left(3\right)\left(2x\right)\left(1\right)\left(2x+1\right)$
By simplifying it, we get
 $=8x^3+1+6x\left(2x+1\right)$
Open the bracket to simplify further
 $\Rightarrow (2x+1{)}^3=8x^3+1+12x^2+6x$
(B) ${\left(2a-3b\right)}^3$
We know that
 ${\left(x-y\right)}^3=x^3-y^3-3xy(x-y)$ . . . . (1)
Here, ${\left(2a-3b\right)}^3$
By comparing the above expression with the formula we have, we can write
 $\Rightarrow x=2a,y=3b$
Now, by substituting the values in the given formula, we get
 $(2a-3b{)}^3={\left(2a\right)}^3-{\left(3b\right)}^3-\left(3\right)\left(2a\right)\left(3b\right)\left(2a-3b\right)$
By simplifying, we get
 $=8a^3-27b^3-18ab\left(2a-3b\right)$
By opening the brackets to further simplify, we get
 $(2a-3b{)}^3=8a^3-27b^3-36a^{2}b+54a{b}^2$
(C) ${\left[{\displaystyle \frac{3}{2}}x+1\right]}^3$
We know that,
 ${\left(x+y\right)}^3=x^3+y^3+3xy(x+y)$
We have ${\left[{\displaystyle \frac{3}{2}}x+1\right]}^3$
By comparing the above expression with the formula we have, we can write
 $x={\displaystyle \frac{3}{2}}x$ and $y=1$
Now, by substituting the values in the given formula, we get
 ${\left[{\displaystyle \frac{3}{2}}x+1\right]}^3={\left[{\displaystyle \frac{3}{2}}x\right]}^3+1^3+3\left[{\displaystyle \frac{3}{2}}x\right]\times 1\times \left[{\displaystyle \frac{3}{2}}x+1\right]$
By simplifying, we get
 $={\displaystyle \frac{27}{8}}{x}^3+1+{\displaystyle \frac{9}{2}}x\left({\displaystyle \frac{3}{2}}x+1\right)$
By opening the brackets to further simplify, we get
 $={\displaystyle \frac{27}{8}}{x}^3+1+{\displaystyle \frac{27}{4}}{x}^2+{\displaystyle \frac{9}{2}}x$
By rearranging it, we get
 ${\left[{\displaystyle \frac{3}{2}}x+1\right]}^3={\displaystyle \frac{27}{8}}{x}^3+{\displaystyle \frac{27}{4}}{x}^2+{\displaystyle \frac{9}{2}}x+1$
(D) ${\left[x-{\displaystyle \frac{2}{3}}y\right]}^3$
We know that
 ${\left(x-y\right)}^3=x^3-y^3-3xy(x-y)$
Here, we have
 ${\left[x-{\displaystyle \frac{2}{3}}y\right]}^3$
By comparing the above expression with the formula we have, we can write
 $x=x$ and $y={\displaystyle \frac{2}{3}}y$
Now, by substituting the values in the given formula, we get
 ${\left[x-{\displaystyle \frac{2}{3}}y\right]}^3=x^3-{\left({\displaystyle \frac{2}{3}}y\right)}^3-3x\left({\displaystyle \frac{2}{3}}y\right)\left(x-{\displaystyle \frac{2}{3}}y\right)$
By simplifying it, we get
 $=x^3-{\displaystyle \frac{8}{27}}{y}^3-2xy\left(x-{\displaystyle \frac{2}{3}}y\right)$
By opening the brackets to further simplify, we get
 $=x^3-{\displaystyle \frac{8}{27}}{y}^3-2x^{2}y+{\displaystyle \frac{4}{3}}x{y}^2$
Therefore, ${\left[x-{\displaystyle \frac{2}{3}}y\right]}^3=x^3-{\displaystyle \frac{8}{27}}{y}^3-2x^{2}y+{\displaystyle \frac{4}{3}}x{y}^2$
Hence, we have written all the cubes in expanded form.

Note: It was a simple question of substituting values in the formula. You need to know the formula to solve such questions. Be careful with positive and negative signs. You can further simplify question (C) and (D) to write the solution into standard form of polynomials. You can do that by multiplying both the sides by the LCM of the denominator of the RHS.