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. •••♪ Reply
Answers:
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.
•••♪
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)