Question

factorise (1) 27a³+64b³ (2) 49y³ - 1000 using the above results​

Answer 1

(7y−10)(49y

2

+70y+100)

343y

3

−1000 = (7y)

3

−(10)

3

We know that x

3

−y

3

=(x−y)(x

2

+xy+y

2

)

Comparing both sides, then we have

x=7y and y=10

Therefore, value of (7y)

3

−(10)

3

is

=(7y−10)[(7y)

2

+(7y)(10)+(10)

2

]

= (7y−10)(49y

2

+70y+100)

Answer 2

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