Factorise: a^2(b^3-c^3)+b^2(c^3-a^3)+c^2(a^3-b^3).........
Answers
Answer:
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Step-by-step explanation:
Equation is alway true
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
a^2*(b^3-c^3)+b^2*(c^3-a^3)+c^2*(a^3-b^3)-((a-b)*(b-c)*(c-a)*(a*b+b*c+c*a))=0
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((((a2)•((b3)-(c3)))+((b2)•((c3)-(a3))))+((c2)•((a3)-(b3))))-(((a-b)•(b-c)•(c-a))•(ab+ac+bc)) = 0
Step 2 :
Equation at the end of step 2 :
((((a2)•((b3)-(c3)))+((b2)•((c3)-(a3))))+((c2)•((a3)-(b3))))-((a-b)•(b-c)•(c-a)•(ab+ac+bc)) = 0
Step 3 :
Equation at the end of step 3 :
((((a2)•((b3)-(c3)))+((b2)•((c3)-(a3))))+((c2)•((a3)-(b3))))-(a-b)•(b-c)•(c-a)•(ab+ac+bc) = 0
Step 4 :
Trying to factor as a Difference of Cubes:
4.1 Factoring: a3-b3
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 : a3 is the cube of a1
Check : b3 is the cube of b1
Factorization is :
(a - b) • (a2 + ab + b2)
Trying to factor a multi variable polynomial :
4.2 Factoring a2 + ab + b2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Equation at the end of step 4 :
((((a2)•((b3)-(c3)))+((b2)•((c3)-(a3))))+c2•(a-b)•(a2+ab+b2))-(a-b)•(b-c)•(c-a)•(ab+ac+bc) = 0
Step 5 :
Trying to factor as a Difference of Cubes:
5.1 Factoring: c3-a3
Check : c3 is the cube of c1
Check : a3 is the cube of a1
Factorization is :
(c - a) • (c2 + ac + a2)
Trying to factor a multi variable polynomial :
5.2 Factoring c2 + ac + a2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Equation at the end of step 5 :
((((a2)•((b3)-(c3)))+b2•(c-a)•(a2+ac+c2))+c2•(a-b)•(a2+ab+b2))-(a-b)•(b-c)•(c-a)•(ab+ac+bc) = 0
Step 6 :
Trying to factor as a Difference of Cubes:
6.1 Factoring: b3-c3
Check : b3 is the cube of b1
Check : c3 is the cube of c1
Factorization is :
(b - c) • (b2 + bc + c2)
Trying to factor a multi variable polynomial :
6.2 Factoring b2 + bc + c2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Equation at the end of step 6 :
((a2•(b-c)•(b2+bc+c2)+b2•(c-a)•(a2+ac+c2))+c2•(a-b)•(a2+ab+b2))-(a-b)•(b-c)•(c-a)•(ab+ac+bc) = 0
Step 7 :
Equation at the end of step 7 :
0 = 0
Step 8 :
Equations which are always true :
8.1 Solve 0 = 0This equation is a tautology (Something which is always true)
Equation is alway true
Step-by-step explanation:
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