Question

factorize p³q³+343/729

Answer 1

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

Here your answer goes

Given ,

$p^3q^3+ \frac{343}{729}$

⇒ We know that 343 , 729 is the cube of 7 and 9 respectively

⇒ $( pq )^3 + ( 7 / 9 )^3$

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

⇒ $( pq +\frac{7}{9} )[ ( pq^2 ) -\frac{7pq}{9} +\frac{49}{81}]$

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