Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(a^{2}-b^{2}\) | (I) | \(a^{2}+b^{2}-2ab\) |
| (B) | \((a-b)^{2}\) | (II) | \((a+b)(a-b)\) |
| (C) | \((a+b)^{3}\) | (III) | \((a^{2}+b^{2}+ab)(a-b)\) |
| (D) | \((a^{3}-b^{3})\) | (IV) | \(a^{3}+b^{3}+3ab(a+b)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
Each expression in List-I is matched with its standard algebraic identity in List-II.
(A) \(a^{2}-b^{2}\) — this matches (II) \((a+b)(a-b)\). The difference of two squares factorises into the sum times the difference: \((a+b)(a-b)=a^{2}-ab+ab-b^{2}=a^{2}-b^{2}\).
(B) \((a-b)^{2}\) — this matches (I) \(a^{2}+b^{2}-2ab\). The square of a difference is the sum of the squares minus twice the product: \((a-b)(a-b)=a^{2}-2ab+b^{2}\).
(C) \((a+b)^{3}\) — this matches (IV) \(a^{3}+b^{3}+3ab(a+b)\). The cube of a sum is \(a^{3}+3a^{2}b+3ab^{2}+b^{3}\), and the two middle terms together are \(3ab(a+b)\).
(D) \((a^{3}-b^{3})\) — this matches (III) \((a^{2}+b^{2}+ab)(a-b)\). The difference of two cubes factorises as \((a-b)(a^{2}+ab+b^{2})\); multiplying out, the terms \(a^{2}b\) and \(ab^{2}\) cancel and leave \(a^{3}-b^{3}\).