Q1.
James and Reshma were talking about algebraic identities they learnt in school. James: (a − b)²(a + b) = (a² − 2ab + b²)(a + b). Reshma: I have a different idea. (a − b)²(a + b) = (a − b)[(a − b)(a + b)] = (a − b)(a² − b²). I will find this product to get the answer. According to you, who is correct and why?
Answer
Both are correct. They are two routes to the same expression, and both routes are legal. Reshma’s is the shorter one.
James’s route — expand the square first:
(a² − 2ab + b²)(a + b)
= a³ + a²b − 2a²b − 2ab² + ab² + b³
= a³ − a²b − ab² + b³
= a³ + a²b − 2a²b − 2ab² + ab² + b³
= a³ − a²b − ab² + b³
Reshma’s route — regroup first, so that a² − b² appears:
(a − b)[(a − b)(a + b)] = (a − b)(a² − b²)
= a³ − ab² − a²b + b³
= a³ − a²b − ab² + b³
= a³ − ab² − a²b + b³
= a³ − a²b − ab² + b³
The two answers are identical, and both factor back as
a³ − a²b − ab² + b³ = a²(a − b) − b²(a − b) = (a − b)(a² − b²) = (a − b)²(a + b)
Why Reshma may regroup: multiplication is associative, so (a − b)²(a + b) may be read as [(a − b)(a − b)](a + b) or as (a − b)[(a − b)(a + b)] — the brackets are ours, not the expression’s. She then spots that one of the pairs, (a − b)(a + b), is a known identity, so she replaces it by a² − b² and is left with a single easy multiplication. James has to multiply a three-term expression by a two-term one, giving six terms, two of which then cancel.
Try This: Combine identities the same way to get new results. For instance (a + b)³(a − b)³ = [(a + b)(a − b)]³ = (a² − b²)³ — three lines of work replaced by one.