Factorise 726p^6-q^6 plz dont give useless answers.
Answers
Answer:
729p6-q6
Final result :
(3p+q)•(9p2-3pq+q2)•(3p-q)•(9p2+3pq+q2)
Step by step solution :
Step 1 :
Equation at the end of step 1 :
36p6 - q6
Step 2 :
Trying to factor as a Difference of Squares :
2.1 Factoring: 729p6-q6
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 : 729 is the square of 27
Check : p6 is the square of p3
Check : q6 is the square of q3
Factorization is : (27p3 + q3) • (27p3 - q3)
Trying to factor as a Sum of Cubes :
2.2 Factoring: 27p3 + q3
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 : 27 is the cube of 3
Check : p3 is the cube of p1
Check : q3 is the cube of q1
Factorization is :
(3p + q) • (9p2 - 3pq + q2)
Trying to factor a multi variable polynomial :
2.3 Factoring 9p2 - 3pq + q2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Trying to factor as a Difference of Cubes:
2.4 Factoring: 27p3 - q3
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 : 27 is the cube of 3
Check : p3 is the cube of p1
Check : q3 is the cube of q1
Factorization is :
(3p - q) • (9p2 + 3pq + q2)
Trying to factor a multi variable polynomial :
2.5 Factoring 9p2 + 3pq + q2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Final result :
(3p+q)•(9p2-3pq+q2)•(3p-q)•(9p2+3pq+q2)
Step-by-step explanation:
please mark me brainliest and follow me