If ∆ is an operation such that for integers a and b we have

a ∆ b = a × b – 2 × a × b + b × b (–a) × b + b × b

then find (i) 4 (– 3)

(ii) (– 7) ∆ (– 1)

Also show that 4 ∆ (– 3) ≠ (– 3) ∆ 4

and (– 7) ∆ (– 1) ≠ ( – 1) ∆ (– 7)

We have been given that,

a ∆ b = a × b – 2 × a × b + b × b (-a) × b + b × b

Simplifying it, we get

a ∆ b = a × b – 2 × a × b – a × b³ + b² …(A)

Apply the same formula in the questions that follows:

(i). We have 4 ∆ (-3).

Put a = 4 and b = -3 in equation (A), we get

4 ∆ (-3) = 4 × (-3) – 2 × 4 × (-3) – 4 × (-3)³ + (-3)²

⇒ 4 ∆ (-3) = -12 + 24 + 108 + 9

⇒ 4 ∆ (-3) = 129

Thus, answer is 129.

(ii). We have (-7) ∆ (-1).

Put a = -7 and b = -1 in equation (A), we get

(-7) ∆ (-1) = (-7) × (-1) – 2 × (-7) × (-1) – (-7) × (-1)³ + (-1)²

⇒ (-7) ∆ (-1) = 7 – 14 – 7 + 1

⇒ (-7) ∆ (-1) = -13

Thus, the answer is -13.

To show: 4 ∆ (-3) ≠ (-3) ∆ 4

LHS: 4 ∆ (-3) = 129 [from answer of part (i)]

RHS: (-3) ∆ 4 = (-3) × 4 – 2 × (-3) × 4 – (-3) × (4)³ + (4)² [by putting a = -3 and b = 4 in equation A]

⇒ (-3) ∆ 4 = -12 + 24 + 192 + 16

⇒ (-3) ∆ 4 = 220

Comparing LHS and RHS, we see that

LHS ≠ RHS [∵ 129 ≠ 220]

To show: (-7) ∆ (-1) ≠ (-1) ∆ (-7)

LHS: (-7) ∆ (-1) = -13 [from answer of part (ii)]

RHS: (-1) ∆ (-7) = (-1) × (-7) – 2 × (-1) × (-7) – (-1) × (-7)³ + (-7)²

⇒ (-1) ∆ (-7) = 7 – 14 – 343 + 49

⇒ (-1) ∆ (-7) = -301

Clearly, LHS ≠ RHS [∵ -13 ≠ -301]