# 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

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