simplify (2p + 5q)^3 - (2p - 5q)^3
Answers
Answer:
2p5q-10pq3/4p3q4+12p5q3
Final result :
p4q • (24pq2 + 4p - 5q6)
————————————————————————
2
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(q3)
(((2•(p5))•q)-(((10p•————)•(p3))•(q4)))+((22•3p5)•q3)
4
Step 2 :
q3
Simplify ——
4
Equation at the end of step 2 :
q3
(((2•(p5))•q)-(((10p•——)•p3)•q4))+(22•3p5q3)
4
Step 3 :
Multiplying exponential expressions :
3.1 p1 multiplied by p3 = p(1 + 3) = p4
Equation at the end of step 3 :
5p4q3
(((2•(p5))•q)-(—————•q4))+(22•3p5q3)
2
Step 4 :
Multiplying exponential expressions :
4.1 q3 multiplied by q4 = q(3 + 4) = q7
Equation at the end of step 4 :
5p4q7
(((2 • (p5)) • q) - —————) + (22•3p5q3)
2
Step 5 :
Equation at the end of step 5 :
5p4q7
((2p5 • q) - —————) + (22•3p5q3)
2
Step 6 :
Rewriting the whole as an Equivalent Fraction :
6.1 Subtracting a fraction from a whole
Rewrite the whole as a fraction using 2 as the denominator :
2p5q 2p5q • 2
2p5q = ———— = ————————
1 2
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Adding fractions that have a common denominator :
6.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
2p5q • 2 - (5p4q7) 4p5q - 5p4q7
—————————————————— = ————————————
2 2
Equation at the end of step 6 :
(4p5q - 5p4q7)
—————————————— + (22•3p5q3)
2
Step 7 :
Rewriting the whole as an Equivalent Fraction :
7.1 Adding a whole to a fraction
Rewrite the whole as a fraction using 2 as the denominator :
(22•3p5q3) (22•3p5q3) • 2
(22•3p5q3) = —————————— = ——————————————
1 2
Step 8 :
Pulling out like terms :
8.1 Pull out like factors :
4p5q - 5p4q7 = p4q • (4p - 5q6)
Trying to factor as a Difference of Squares :
8.2 Factoring: 4p - 5q6
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 : 4 is the square of 2
Check : 5 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares.
Trying to factor as a Difference of Cubes:
8.3 Factoring: 4p - 5q6
Theory : A difference 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 : 4 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Adding fractions that have a common denominator :
8.4 Adding up the two equivalent fractions
p4q • (4p-5q6) + (22•3p5q3) • 2 24p5q3 + 4p5q - 5p4q7
——————————————————————————————— = —————————————————————
2 2
Step 9 :
Pulling out like terms :
9.1 Pull out like factors :
24p5q3 + 4p5q - 5p4q7 = p4q • (24pq2 + 4p - 5q6)
Trying to factor a multi variable polynomial :
9.2 Factoring 24pq2 + 4p - 5q6
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Final result :
p4q • (24pq2 + 4p - 5q6)
————————————————————————
2
Step-by-step explanation:
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