Question

factorise 24a^4+81a by sum of different cubes​

Answer 1

Step-by-step explanation:

STEP

1:

Equation at the end of step 1

(24 • (a3)) + 34b3

STEP 2 :

Equation at the end of step2:

(23•3a3) + 34b3

STEP3:

STEP4:

Pulling out like terms

4.1 Pull out like factors :

24a3 + 81b3 = 3 • (8a3 + 27b3)

Trying to factor as a Sum of Cubes:

4.2 Factoring: 8a3 + 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 : 8 is the cube of 2

Check : 27 is the cube of 3

Check : a3 is the cube of a1

Check : b3 is the cube of b1

Factorization is :

(2a + 3b) • (4a2 - 6ab + 9b2)

Trying to factor a multi variable polynomial :

4.3 Factoring 4a2 - 6ab + 9b2

Try to factor this multi-variable trinomial using trial and error

Factorization fails

Final result :

3 • (2a + 3b) • (4a2 - 6ab + 9b2)

Answer 2

STEP

1:

Equation at the end of step 1

(24 • (a3)) + 34b3

STEP 2 :

Equation at the end of step2:

(23•3a3) + 34b3

STEP3:

STEP4:

Pulling out like terms

4.1 Pull out like factors :

24a3 + 81b3 = 3 • (8a3 + 27b3)

Trying to factor as a Sum of Cubes:

4.2 Factoring: 8a3 + 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 : 8 is the cube of 2

Check : 27 is the cube of 3

Check : a3 is the cube of a1

Check : b3 is the cube of b1

Factorization is :

(2a + 3b) • (4a2 - 6ab + 9b2)

Trying to factor a multi variable polynomial :

4.3 Factoring 4a2 - 6ab + 9b2

Try to factor this multi-variable trinomial using trial and error

Factorization fails

Final result :

3 • (2a + 3b) • (4a2 - 6ab + 9b2)