Question

Factorise: a6+27b3 please answer fasttttttttttttt

Answer 1

Changes made to your input should not affect the solution:

(1): "b3" was replaced by "b^3". 1 more similar replacement(s).

Step by step solution :

Step 1 :

Equation at the end of step 1 :

(a6) + 33b3

Step 2 :

Trying to factor as a Sum of Cubes :

2.1 Factoring: a6+27b3

Theory : A sum of two perfect cubes, a3 + b3 can be factored into :

(a+b) • (a2-ab+b2)

Proof : (a+b) • (a2-ab+b2) =

a3-a2b+ab2+ba2-b2a+b3 =

a3+(a2b-ba2)+(ab2-b2a)+b3=

a3+0+0+b3=

a3+b3

Check : 27 is the cube of 3

Check : a6 is the cube of a2

Check : b3 is the cube of b1

Factorization is :

(a2 + 3b) • (a4 - 3a2b + 9b2)

this is ur answer I hope u will understand.

Answer 2

$= ( {a}^{6} + 27 {b}^{3} )$

$= ( {a}^{2} {)}^{3} + (3b {)}^{3}$

$= ( {a}^{2} + 3b)( {a}^{4} - 3 {a}^{2}b \: + 9 {b}^{2} )$