Question

(a3+b3)=(a-b)(a2+b2+ab)

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Answer 1

Why is a³+b³= (a+b) (a²-ab+b²) if we don't have any negative sign?

We all know,

(a+b)³ = a³ + 3a²b + 3ab² + b³

Now solve the above equation to get the required equation,

or, a³+b³ = (a+b)³ - 3a²b - 3ab²

or, a³+b³ = (a+b)³ - 3ab(a + b)

or, a³+b³ = (a+b) {(a+b)² - 3ab}

or, a³+b³ = (a+b) {a² + 2ab + b² - 3ab}

or, a³+b³ = (a+b) {a² - ab + b²}

Let's check:

a³+b³ = (a+b) {a² - ab + b²}

Now solve only R. H. S, we get

(a+b) {a² - ab + b²}

= a³ - a²b + ab² + a²b - ab² + b³

cancel - a²b + a²b and + ab² - ab²

= a³+b³ = L. H. S (proved)

i.e. a³+b³ = (a+b) {a² - ab + b²}

If we do not have any negative sign then what will happen:

i.e. instead of using “-ab” we used “+ab”.

a³+b³ = (a+b) {a² + ab + b²}

Now solve R. H. S, we get

(a+b) {a² + ab + b²}

= a³ + a²b + ab² + a²b + ab² + b³

= a³ + 2a²b + 2ab² + b³

= a³+ b³ + 2ab(a+b) ≠ L. H. S

i.e. a³+b³ ≠ (a+b) {a² + ab + b²}

Negative sign is required to fulfill the above contention i.e. a³+b³=(a+b){a² - ab + b²}

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