Question

Factorise using identities. i) 8p^3 – 27q^3+r^3+18pqr

Answer 1

Answer:

STEP

1

:

Equation at the end of step 1

(((8 • (p3)) - 33q3) + r3) + 18pqr

STEP

2

:

Equation at the end of step

2

:

((23p3 - 33q3) + r3) + 18pqr

STEP

3

:

Checking for a perfect cube

3.1 8p3+18pqr-27q3+r3 is not a perfect cube

Final result :

8p3 + 18pqr - 27q3 + r3

Answer 2

Step-by-step explanation:

${8p}^{3} - {27q}^{3} + {r}^{3} +18pqr$

$= {(2p)}^{3} + {( - 3q)}^{3} + {(r)}^{3} - 3(2p)( - 3q)(r)$

Using the identity

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$

Here, a=2p, b=-3q, c=r

$=(2p + ( - 3q) + r) ({(2p)}^{2} + {( - 3q)}^{2} + {(r)}^{2} - (2p)( - 3q) - ( - 3q)(r) - (r)(2p))$

$= (2p - 3q + r)( {4p}^{2} + {9q}^{2} + {r}^{2} + 6pq - 3qr - 2rp$

hope this helps:D

P.S.- This might be kinda confusing to look, but after solving you'll know how easy it is:p