Question

factorise
p^3q^3+343/729

Answer 1

Hello Mate!

p^3q^3 + 343 / 729

Here, 343 = 7 × 7 × 7 = 7^3
729 = 9 × 9 × 9 = 9^3

p^3q^3 + 7^3 / 9^3

( pq )^3 + ( 7 / 9 )^3

a^3 + b^3 = ( a + b )( a^2 - ab + b^2 )

= ( pq + 7/9 )[ ( pq^2 ) - 7pq/9 + 49/81 ]

Hope it helps☺!✌

Answer 2

Answer :

Now, p³q³ + 343/729

= (pq)³ + (7/9)³

= (pq + 7/9) {(pq)² - (pq × 7/9) + (7/9)²},

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

= (pq + 7/9) (p²q² - 7pq/9 + 49/81),

which is the required factorization.

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