Question

Verify: (i) x3 + y3 = (x + y) (x2 – xy + y2) (ii) x3 - y3 = (x - y) (x2 + xy + y2)​

Answer 1

Answer:

let x =2 y =4

in 1st case

2^3+4^3 = (2+4)(2^2-2(4)+4^2)

8+64 = 6(4-8+16)

72=6(12)

72=72

in 2nd case

2^3-4^3 = (2-4)(2^2+2(4)+4^2)

8-64 = (-2)(4+8+16)

-56 = (-2)(28)

-56 = -56

Answer 2

(I) x³+y³

We know that,

(x+y)³ = x³+y³+3xy(x+y)

Hence,

x³+y³=(x+y)³ - 3xy(x+y)

Let x + y = z

Hence,

x³+y³ = z³ - 3xyz

Factor out common z

→ x³+y³ = z(z²-3xy)

Substitute the value of z

→x³+y³ = (x+y)[(x+y)²-3xy)]

We know that,

(x+y)² = x²+y²+2xy

Hence,

→ x³+y³ = (x+y)(x²+y²+2xy-3xy)

→ x³+y³ = (x+y)(x²+y²-xy)

Hence, verified.

(ii)

We know that

(x-y)³=x³-y³-3xy(x-y)

Hence,

→ x³-y³ = (x-y)³ + 3xy(x-y)

→ x³-y³ = (x-y)[(x-y)² + 3xy]

Since, (x-y)² = x²+y²-2xy

Hence,

→ x³-y³ = (x-y)(x²+y²-2xy+3xy)

→ x³-y³ = (x-y)(x²+y²+xy)

Hence, verified.