Question

factorise 27p^3q^3+64r^3 ​

Answer 1

Step-by-step explanation:

27p³q³ + 64r³

(3pq)³ + (4r)³

(3pq + 4r) { (3pq)² - (3pq)(4r) + (4r)² }

(3pq + 4r) { 9p²q² - 12pqr + 16r² }

Answer 2

$\huge{\pink{\overbrace{\underbrace{\purple{\mathfrak{ANSWER}}}}}}$

$27 {p}^{3} {q}^{3} + 64 {r}^{3} \\ {(3pq)}^{3} + {(4r)}^{3} \\ (3pq + 4r)( {(3pq)}^{2} - 3pq \times 4r + {(4r)}^{2} ) \\ (3pq + 4r)(9 {p}^{2} {q}^{2} - 12pqr + 16 {r}^{2} )$

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