Assertion (A): If a = 2, b = -6 and c = 4, then $\dfrac{3abc - a^3 - b^3 - c^3}{ab + bc + ca - a^2 - b^2 - c^2}= 0$

Reason (R): If a + b + c = 0 then a³ + b³ + c³ + 3abc = 0

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

A is true, R is false.

Explanation

Given, a = 2, b = -6 and c = 4,

L.H.S.

$= \dfrac{3abc - a^3 - b^3 - c^3}{ab + bc + ca - a^2 - b^2 - c^2}\\[1em] = \dfrac{3 \times 2 \times (-6) \times 4 - 2^3 - (-6)^3 - 4^3}{2 \times (-6) + (-6) \times 4 + 4 \times 2 - 2^2 - (-6)^2 - 4^2}\\[1em] = \dfrac{-144 - 8 + 216 - 64}{-12 - 24 + 8 - 4 - 36 - 16}\\[1em] = 0$

R.H.S. = 0

∴ L.H.S. = R.H.S.

∴ Assertion(A) is true.

Given,

a + b + c = 0

⇒ a + b = -c

Cubing both side, we get

⇒ (a + b)³ = (-c)³

⇒ a³ + b³ + 3ab(a + b) = -c³

⇒ a³ + b³ + 3ab(-c) = -c³

⇒ a³ + b³ - 3abc = -c³

⇒ a³ + b³ + c³ - 3abc = 0

∴ Reason(R) is false.

Hence, Assertion (A) is true, Reason (R) is false.