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Answers
Explanation:
òlChanges made to your input should not affect the solution:
(1): "a3" was replaced by "a^3".
STEP
1
:
Equation at the end of step 1
(26•3a3) + 3
STEP
2
:
STEP
3
:
Pulling out like terms
3.1 Pull out like factors :
192a3 + 3 = 3 • (64a3 + 1)
Trying to factor as a Sum of Cubes:
3.2 Factoring: 64a3 + 1
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 : 64 is the cube of 4
Check : 1 is the cube of 1
Check : a3 is the cube of a1
Factorization is :
(4a + 1) • (16a2 - 4a + 1)
Trying to factor by splitting the middle term
3.3 Factoring 16a2 - 4a + 1
The first term is, 16a2 its coefficient is 16 .
The middle term is, -4a its coefficient is -4 .
The last term, "the constant", is +1
Step-1 : Multiply the coefficient of the first term by the constant 16 • 1 = 16
Step-2 : Find two factors of 16 whose sum equals the coefficient of the middle term, which is -4 .
-16 + -1 = -17
-8 + -2 = -10
-4 + -4 = -8
-2 + -8 = -10
-1 + -16 = -17
1 + 16 = 17
2 + 8 = 10
4 + 4 = 8
8 + 2 = 10
16 + 1 = 17
Observation : No two such factors can be found !!