The algebraic identity in List-I and List-II are given below. Match List-I with List-II for correct relation between them.
| List-I | List-II | ||
|---|---|---|---|
| (A) | \((a+b)^{2}\) | (I) | \(a^{2}+b^{2}-2ab\) |
| (B) | \((a-b)^{2}\) | (II) | \(a^{2}+b^{2}+2ab\) |
| (C) | \((a+b)^{3}\) | (III) | \(a^{3}-b^{3}-3ab(a-b)\) |
| (D) | \((a-b)^{3}\) | (IV) | \(a^{3}+b^{3}+3ab(a+b)\) |
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) – (II), (B) – (I), (C) – (IV), (D) – (III)
Each identity comes from multiplying out the bracket. With \((a-b)\) the terms carrying an odd power of b take a minus sign.
(A) \((a+b)^{2}\) — this matches (II) \(a^{2}+b^{2}+2ab\). \((a+b)(a+b)=a^{2}+2ab+b^{2}\).
(B) \((a-b)^{2}\) — this matches (I) \(a^{2}+b^{2}-2ab\). The middle term \(2ab\) contains b once, so it turns negative.
(C) \((a+b)^{3}\) — this matches (IV) \(a^{3}+b^{3}+3ab(a+b)\). Multiplying out gives \(a^{3}+3a^{2}b+3ab^{2}+b^{3}\), and the two middle terms together are \(3ab(a+b)\).
(D) \((a-b)^{3}\) — this matches (III) \(a^{3}-b^{3}-3ab(a-b)\). It equals \(a^{3}-3a^{2}b+3ab^{2}-b^{3}\), and the middle terms together are \(-3ab(a-b)\).