Factorization involves breaking down an expression.
The sum of cubes is given as:
[tex](a^3 + b^3) = (a + b)(a^2 -ab + b^2)[/tex]
(a) Verify the formula
Expand the expression on the right-hand side
[tex](a^3 + b^3) = a^3 -a^2b + ab^2 +a^2b - ab^2 + b^3[/tex]
Collect like terms
[tex](a^3 + b^3) = a^3 -a^2b +a^2b+ ab^2 - ab^2 + b^3[/tex]
[tex](a^3 + b^3) = a^3 + b^3[/tex]
The formula has been verified
(b) Factorized 8x^3+ 27
We have:
[tex]8x^3 + 27[/tex]
Express 27 as 3^3
[tex]8x^3 + 27 =8x^3 + 3^2[/tex]
Express 8 as 2^3
[tex]8x^3 + 27 =2^3x^3 + 3^3[/tex]
Rewrite as:
[tex]8x^3 + 27 =(2x)^3 + 3^3[/tex]
Given that:
[tex](a^3 + b^3) = (a + b)(a^2 -ab + b^2)[/tex]
The expression becomes
[tex]8x^3 + 27 =(2x + 3)((2x)^2 - (2x)3 +3^2)[/tex]
[tex]8x^3 + 27 =(2x + 3)(4x^2 -8x3 +9)[/tex]
(c)The other factor of 2^3 - b^3
By difference of cubes, we have:
[tex]x^3 - y^3=(x-y)(x^2+xy+y^2)[/tex]
So, the equation becomes
[tex]2^3 - b^3=(2-b)(2^2+2b+b^2)[/tex]
This gives
[tex]2^3 - b^3=(2-b)(4+2b+b^2)[/tex]
Hence, the other factor of [tex]2^3 - b^3[/tex] is [tex]4+2b+b^2[/tex]
(c) Factor x^3 - 1
We have:
[tex]x^3 - y^3=(x-y)(x^2+xy+y^2)[/tex]
Express 1 as 1^3 in x^3 - 1
[tex]x^3 - 1 =x^3 - 1^3[/tex]
[tex]x^3 - y^3=(x-y)(x^2+xy+y^2)[/tex] becomes
[tex]x^3 - 1^3=(x-1)(x^2+x+1^2)[/tex]
[tex]x^3 - 1^3=(x-1)(x^2+x+1)[/tex]
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