Question

7. Verify the following: i) (ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0 ii) (a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³+ c³ – 3abc iii) (p – q) (p² + pq + q²) = p³ – q³. EXPLAIN

Answer 1

1. (ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0

We know that (a+b)(a-b) = a²-b²

(ab + bc)(ab -bc) can be written as a²b² - b²c²

(bc + ca)(bc -ca) can be written as b²c² - c²a²

(ca + ab)(ca - ab) can be written as c²a² - a²b²

→ a²b² - b²c² + b²c² - c²a² + c²a² - a²b²

→ a²b² - a²b² - b²c² + b²c² - c²a² + c²a²

→ 0

2. (a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³+ c³ – 3abc

→ a³ + ab² + ac² -a²b - abc -ca² + a²b + b³ + bc² - ab² - b²c - abc + a²c + b²c + c³ - abc - bc² - c²a

→ a³ + b³+ c³ + (- abc - abc - abc) + (ab² - ab² )+ (ac² - ca² ) -(a²b + a²b )+ (bc² - bc² )+ (a²c - c²a) + (b²c - b²c)

→ a³+b³+c³ - 3 abc .

3. (p – q) (p² + pq + q²) = p³ – q³.

→ p³ + p²q + pq² - p²q - pq² - q³

→ p³ - q³ +(p²q - p²q) + (pq² - pq²)

→ p³ - q³