factories this.
x^6 +y^3
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Step by step solution :
Step 1 :
Trying to factor as a Difference of Squares :
1.1 Factoring: x6-y3
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : x6 is the square of x3
Check : y3 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares
Step 1 :
Trying to factor as a Difference of Squares :
1.1 Factoring: x6-y3
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : x6 is the square of x3
Check : y3 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares
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