SUBTRACT a3+b3+3abc
Answers
Answer:
a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)
Equation is alway true
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "c2" was replaced by "c^2". 5 more similar replacement(s).
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
a^3+b^3+c^3-3*a*b*c-((a+b+c)*(a^2+b^2+c^2-a*b-b*c-c*a))=0
Step by step solution :
Step 1 :
Trying to factor by pulling out :
1.1 Factoring: a2-ab-ac+b2-bc+c2
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -ac-bc
Group 2: a2-ab
Group 3: b2+c2
Pull out from each group separately :
Group 1: (a+b) • (-c)
Group 2: (a-b) • (a)
Group 3: (b2+c2) • (1)
Looking for common sub-expressions :
Group 1: (a+b) • (-c)
Group 3: (b2+c2) • (1)
Group 2: (a-b) • (a)
Bad news !! Factoring by pulling out fails :
The groups have no common factor and can not be added up to form a multiplication.
Equation at the end of step 1 :
((((a3)+(b3))+(c3))-3abc)-(a+b+c)•(a2-ab-ac+b2-bc+c2) = 0
Step 2 :
Equation at the end of step 2 :
0 = 0
Step 3 :
Equations which are always true :
3.1 Solve 0 = 0This equation is a tautology (Something which is always true)
Equation is alway true
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