find the formula for a³-b³ & a³+b³ by algebraic calculation.
Answers
Step-by-step explanation:
a^3+b^3=(a+b)(a^2-ab+b^2)
a^3-b^3=(a-b)(a^2+ab+b^2)
Answers:
- a³ - b³ = (a - b)(a² + ab + b²)
- a³ + b³ = (a + b)(a² - ab + b²)
Step By Step Solution:
We have to show - a³ - b³ = (a - b)(a² + ab + b²)
We know that,
→ (a - b)³ = a³ - 3a²b + 3ab² - b³
(a - b)³ can be written as,
= (a - b)² × (a - b)
= (a - b)(a² - 2ab + b²)
So,
→ (a - b)(a² - 2ab + b²) = a³ - b³ - 3ab(a - b)
Moving -3ab(a - b) to the left side, we get,
→ (a - b)(a² - 2ab + b²) + 3ab(a - b) = a³ - b³
Taking (a - b) as common, we get,
→ a³ - b³ = (a - b)(a² - 2ab + b² + 3ab)
→ a³ - b³ = (a - b)(a² + ab + b²) which is the required formula.
We have to show - a³ + b³ = (a + b)(a² - ab + b²)
We know that,
→ (a + b)³ = a³ + b³ + 3ab(a + b)
On transposing, we get,
→ a³ + b³ = (a + b)³ - 3ab(a + b)
Taking (a + b) as common, we get,
→ a³ + b³ = (a + b)[(a + b)² - 3ab]
→ a³ + b³ = (a + b)[a² + 2ab + b² - 3ab]
→ a³ + b³ = (a + b)(a² - ab + b² - 3ab) which is our required formula.
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