Question

Hey guys

Solve the attachment 8 questions using this identity :

(a - b)³= a³ - 3a²b + 3ab² - b²​

Answer 1

1. (2m - 5)³

For this question we will use this identity ⇒ (a - b)³= a³ - 3a²b + 3ab² - b³

Here,

a = 2m

b = 5

So let's expand.

⇒ (2m - 5)³

⇒ (2m)³ - (3) (2m)² (5) + (3) (2m) (5²) - 5³

⇒ 8m³ - 3 (4m²) (5) + (3) (2m) (25) - 125

⇒ 8m³ - 15 (4m²) + 75 (2m) - 125

⇒ 8m³ - 60m² + 150m - 125

$\rule{300}{1}$

2. (4 - p)³

For this question we will use this identity ⇒ (a - b)³= a³ - 3a²b + 3ab² - b³

Here,

a = 4

b = p

So let's expand.

⇒ 4³ - (3) (4²) (p) + (3) (4) (p²) - p³

⇒ 64 - 3 (16) (p) + (12) (p²) - p³

⇒ 64 - 48p + 12p² - p³

$\rule{300}{1}$

3. (7x - 9y)³

For this question we will use this identity ⇒ (a - b)³= a³ - 3a²b + 3ab² - b³

Here,

a = 7x

b = 9y

So let's expand.

⇒ (7x)³ - (3) (7x)² (9y) + (3) (7x) (9y)² - (9y)³

⇒ 343³ - 3 (49x²) (9y) + 3 (7x) (81y²) - 729y³

⇒ 343³ - 147x² (9y) + 21x (81y²) - 729y³

⇒ 343³ - 1323x²y + 1701xy² - 729y³

$\rule{300}{1}$

4. (58)³

For this question we will use this identity ⇒ (a - b)³= a³ - 3a²b + 3ab² - b³

Here,

a = 60

b = 2

So let's expand.

⇒ 60³ - (3) (60²) (2) + (3) (60) (2²) - 2³

⇒ 216000 - 3 (3600) (2) + 180 (4) - 8

⇒ 216000 - 6 (3600) + 720 - 8

⇒ 216000 - 21600 + 720 - 8

⇒ 216000 + 720 - 21600 - 8

⇒ 216720 - 21600 - 8

⇒ 195120 - 8

⇒ 195112

$\rule{300}{1}$

5. (198)³

For this question we will use this identity ⇒ (a - b)³= a³ - 3a²b + 3ab² - b³

Here,

a = 200

b = 2

So let's expand.

⇒ 200³ - (3) (200²) (2) + (3) (200) (2²) - 2³

⇒ 8000000 - 6 (40000) + 12 (200) - 8

⇒ 8000000 - 240000 + 2400 - 8

⇒ 8000000 + 2400 - 240000 - 8

⇒ 8002400 - 240000 - 8

⇒ 7762400 - 8

⇒ 7762392

$\rule{300}{1}$

6. $\sf (2p - \frac{1}{2p})^{3}$

For this question we will use this identity ⇒ (a - b)³= a³ - 3a²b + 3ab² - b³

Here,

a = 2p

b = $\sf \frac{1}{2p}$

So let's expand.

⇒ $\sf (2p)^{3} - (3) \ (2p)^{2} \ (\frac{1}{2p}) + (3) \ (2p) \ (\frac{1}{2p})^{2} -(\frac{1}{2p})^{3}$

⇒ $\sf 8p^{3} - 3(2p)(2p)(\frac{1}{2p})+3(2p)(\frac{1}{2p})(\frac{1}{2p})-\frac{1}{8^{3}}$

⇒ $\sf 8p^{3} - 3(2p)+3(\frac{1}{2p})-\frac{1}{8^{3}}$

⇒ $\sf 8p^{3} - 6p+\frac{3}{2p}-\frac{1}{8^{3}}$

$\rule{300}{1}$

7. $\sf (1-\frac{1}{a})^{3}$

For this question we will use this identity ⇒ (a - b)³= a³ - 3a²b + 3ab² - b³

Here,

a = 1

b = $\sf \frac{1}{a}$

So let's expand.

⇒ $\sf 1^{3}-(3)(1^{2})(\frac{1}{a})+(3)(1)(\frac{1}{a})^{2} - {\frac{1}{a})^{3}$

⇒ $\sf 1-(3)(1)(\frac{1}{a})+(3)(1)(\frac{1}{a^{2}})- {\frac{1}{a^{3}}$

⇒ $\sf 1-3(\frac{1}{a})+3(\frac{1}{a^{2}})- {\frac{1}{a^{3}}$

⇒ $\sf 1-\frac{3}{a}+\frac{3}{a^{2}}- {\frac{1}{a^{3}}$

$\rule{300}{1}$

8. $\sf (\frac{x}{3}- \frac{3}{x})^{3}$

For this question we will use this identity ⇒ (a - b)³= a³ - 3a²b + 3ab² - b³

Here,

a = $\sf \frac{x}{3}$

b = $\sf \frac{3}{x}$

So let's expand.

⇒ $\sf (\frac{x}{3})^{3} - (3)(\frac{x}{3})^{2}(\frac{3}{x})+(3)(\frac{x}{3})(\frac{3}{x})^{2} -( \frac{3}{x})^{3}$

⇒ $\sf \frac{x^{3}}{27} - (3)(\frac{x}{3})(\frac{x}{3}) (\frac{3}{x})+(3)(\frac{x}{3})(\frac{3}{x})(\frac{3}{x}) -\frac{27}{x^{3}}$

⇒ $\sf \frac{x^{3}}{27} - (3)(\frac{x}{3})+(3)(\frac{3}{x}) -\frac{27}{x^{3}}$

⇒ $\sf \frac{x^{3}}{27} -x+\frac{9}{x} -\frac{27}{x^{3}}$

$\rule{300}{1}$

Answer 2

Answer:

$\huge{\underline{\mathtt{\red{❥A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$Expressed in words, the difference of the cubes of two quantities is the product of the difference of the two quantities by the “imperfect square of the sum.”

Proof:

We know the well-known formula

(a-b)³=a³-3 a²b+3 ab²-b³

By transposition,

a³ - b³ = (a-b)³ + 3 a²b - 3 ab²

a³ - b³ = (a-b)³ +3 ab(a-b)

a³ - b³ = (a-b) [(a-b)² +3 ab]

a³ - b³ = (a-b) [(a-b)² +3 ab]

We all know (a - b)² = a² - 2 ab + b²

So

a³ - b³ = (a-b) [(a² - 2 ab + b²) +3 ab]

a³-b³= (a-b)(a²+ab+ b²) [Proved]