Problems on Expanding of (a ± b)$^{3}$ and its Corollaries

Here we will solve different types of application problems on expanding of (a ± b)$^{3}$ and its corollaries.

1. Expanding the following:

(i) (1 + x)$^{3}$

(ii) (2a – 3b)$^{3}$

(iii) (x + $\frac{1}{x}$)$^{3}$

Solution:

(i) (1 + x)$^{3}$ = 1$^{3}$ + 3 ∙ 1$^{2}$ ∙ x + 3 ∙ 1 ∙ x$^{2}$ + x$^{3}$

                 = 1 + 3x + 3x$^{2}$ + x$^{3}$

(ii) (2a – 3b)$^{3}$ = (2a)$^{3}$ - 3 ∙ (2a)$^{2}$ ∙ (3b) + 3 ∙ (2a) ∙ (3b)$^{2}$ – (3b)$^{3}$

                     = 8a$^{3}$ – 36a$^{2}$b + 54ab$^{2}$ – 27b$^{3}$

(iii) (x + $\frac{1}{x}$)$^{3}$ = x$^{3}$ + 3 ∙ x$^{2}$ ∙ $\frac{1}{x}$ + 3 ∙ x ∙ $\frac{1}{x^{2}}$ + $\frac{1}{x^{3}}$

                    = x$^{3}$ + 3x + $\frac{3}{x}$ + $\frac{1}{x^{3}}$.

2. Simplify: $(\frac{x}{2} + \frac{y}{3})^{3} - (\frac{x}{2} - \frac{y}{3})^{3}$

Solution:

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

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

= $2\left \{3 \cdot \frac{x^{2}}{4} \cdot \frac{y}{3} + \frac{y^{3}}{27}\right\}$

= $\frac{x^{2}y}{2} + \frac{2y^{3}}{27}$.

3. Express 8a$^{3}$ – 36a$^{2}$b + 54ab$^{2}$ – 27b$^{3}$ as a perfect cube and find its value when a = 3, b = 2.

Solution:

Given expression = (2a)$^{3}$ – 3(2a)$^{2}$ ∙ 3b + 3  ∙ (2a) ∙ (3b)$^{2}$ – (3b)$^{3}$

                         = (2a – 3b)$^{3}$

When a = 3 and b = 2, the value of the expression = (2 × 3 – 3 × 2)$^{3}$

                                                                          = (6 – 6)$^{3}$

                                                                          = (0)$^{3}$

                                                                          = 0.

4. If x + y = 6 and x$^{3}$ + y$^{3}$ = 72, find xy.

Solution:

We know that (a + b)$^{3}$ – (a$^{3}$ + b$^{3}$) = 3ab(a + b).

Therefore, 3xy(x + y) = (x + y)$^{3}$ – (x$^{3}$ + y$^{3}$)

Or, 3xy ∙ 6 = 6$^{3}$ – 72

Or, 18xy = 216 – 72

Or, 18xy = 144

Or, xy = $\frac{1}{18}$ ∙ 144

Therefore, xy = 8

5. Find a$^{3}$ + b$^{3}$ if a + b = 5 and ab = 6.

Solution:

We know that a$^{3}$  + b$^{3}$ = (a + b)$^{3}$ - 3ab(a + b).

Therefore, a$^{3}$  + b$^{3}$ = 5$^{3}$ – 3 ∙ 6 ∙ 5

                            = 125 – 90

                            = 35.

6. Find x$^{3}$ - y$^{3}$   if x – y = 7and xy = 2.

Solution:

We know that a$^{3}$  - b$^{3}$ = (a - b)$^{3}$ + 3ab(a - b).

Therefore, x$^{3}$  - y$^{3}$ = (x - y)$^{3}$ + 3xy(x - y)

                           = (-7)$^{3}$ + 3 ∙ 2 ∙ (-7)

                           = - 343 – 42

                           = -385.

7. If a - $\frac{1}{a}$ = 5, find a$^{3}$   - $\frac{1}{a^{3}}$.

Solution:

a$^{3}$   - $\frac{1}{a^{3}}$ = (a - $\frac{1}{a}$)$^{3}$ + 3 ∙ a ∙ $\frac{1}{a}$(a - $\frac{1}{a}$)

             = 5$^{3}$ + 3 ∙ 1 ∙ 5

             = 125 + 15

             = 140.

8. If x$^{2}$ + $\frac{1}{a^{2}}$ = 7, find x$^{3}$   + $\frac{1}{x^{3}}$.

Solution:

We know, (x + $\frac{1}{x}$)$^{2}$ = x$^{2}$ + 2 ∙ x ∙ $\frac{1}{x}$ + $\frac{1}{x^{2}}$

                             = x$^{2}$ + $\frac{1}{x^{2}}$ + 2

                             = 7 + 2

                             = 9.

Therefore, x + $\frac{1}{x}$ = $\sqrt{9}$ = ±3.

Now, x$^{3}$ + $\frac{1}{x^{3}}$ = (x + $\frac{1}{x}$)$^{3}$ - 3 ∙ x ∙ $\frac{1}{x}$(x + $\frac{1}{x}$)

                     = (x + $\frac{1}{x}$)$^{3}$ - 3(x + $\frac{1}{x}$).

If x + $\frac{1}{x}$ = 3, x$^{3}$ + $\frac{1}{x^{3}}$ = 3$^{3}$ - 3 ∙ 3

                                  = 27 – 9

                                  = 18.

If x + $\frac{1}{x}$ = -3, x$^{3}$ + $\frac{1}{x^{3}}$ = (-3)$^{3}$ - 3 ∙ (-3)

                                  = -27 + 9

                                  = -18.

9th Grade Math

From Problems on Expanding of (a ± b)$^{3}$ and its Corollaries to HOME PAGE