# a3+b3+c3-3abc=

(a+b+c) X (a2+b2+c2-ab-bc-ca)

• 28

(a+b+c) X (a2+b2+c2-ab-bc-ca)

• 88

above ans is correct

• -11

Moreover if a+b+c = 0

then a3+ b3 + c3 = 3abc

• 11

• -14
1. Factorise x4+x2y2+y4
• -9

hey take it easy mene mazak me hi lia sb kch.. n j mene kha vo bh mazak hi tha.. !

• -10

1/2(a+b+c)[(a-b)²(b-c)²(c-a)²]

• -13

• -10
It do not give any result.....Moreover, it comes ZERO when a+b+c = 0
• -26
(a + b + c)(a2 + b2 + c2 –ab –bc –ca)
• 2
woaaaaahhh!!!!amazing andsuer

• -5
(a+b+c)(a2+b2+c2-ab-bc-ca)
• -8
(a+b+c)(a?+b?+c?-ab-bc-ca)
• -6
(a+b+c) X (a2+b2+c2-ab-bc-ca
• 4
(a+b+c)(a2+b2+c2-ab-bc-ca)
• 12
Baccho vale ques. mat ask karo
• -8
a+b+c
• -8
Taking RHS of the identity: (a + b + c)(a2 + b2 + c2 - ab - bc - ca ) Multiply each term of first polynomial with every term of second polynomial, as shown below: = a(a2 + b2 + c2 - ab - bc - ca ) + b(a2 + b2 + c2 - ab - bc - ca ) + c(a2 + b2 + c2 - ab - bc - ca ) = { (a X a2) + (a X b2) + (a X c2) - (a X ab) - (a X bc) - (a X ca) } + {(b X a2) + (b X b2) + (b X c2) - (b X ab) - (b X bc) - (b X ca)} + {(c X a2) + (c X b2) + (c X c2) - (c X ab) - (c X bc) - (c X ca)} Solve multiplication in curly braces and we get: = a3 + ab2 + ac2 – a2b - abc - a2c + a2b + b3 + bc2 - ab2 – b2c - abc + a2c + b2c + c3 - abc – bc2 - ac2 Rearrange the terms and we get: = a3 + b3 + c3 + a2b – a2b + ac2- ac2 + ab2 - ab2 + bc2 – bc2 + a2c - a2c + b2c – b2c - abc - abc - abc Above highlighted like terms will be subtracted and we get: = a3 + b3 + c3 - abc - abc - abc Join like terms i.e (-abc) and we get: = a3 + b3 + c3 – 3abc Hence, in this way we obtain the identity i.e. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
• -5
a+b+ c=0
• -5
(a+b+c)(a2+b2+c2-ab-bc-ca)
• -4 • 8
The photo shows the solution to this answer for both the cases • 42
achaa
hai...
• -1
(a+b+c)(a2+b+c2-ab-bc-ca)
• -7
(a+b+c) (a2+b2+c2-ab-bc-ca)

• -4
(a+b+c)(a2+b2+c2-ab-bc-ca
• -4
(a+b+c)(a2+b2+c2-ab-bc-ca) is the 8 and the last identity till date in algebraic xpression.
• -2
(a+b+c)(a2+b2+c2-ab-bc-ca)
• -4
men nu ki pata jyada mat chodiyo
• -1

• -2
hi,
(a+b+c)(a^2 +b^2+c^2 -ab-ac-bc)
if a+b+c=0 then your question =0
hope it helps!!
• -5
(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
• -5
correct

• 1
(a+b+c)(a2+b2+c2-ab-bc-ca)

Hope this helps :) ATB!!!!!!

• 8
The answer for the identity a3 + b3 + c3 - 3abc =  (a + b+ c ) ( a2 + b2 + c2 - ab - bc - ac )

However if your answer where a + b+ c = 0, then a​3 + b3 + c3 = 3abc

Sincerely hoping that this answer helps.
• 6
(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
• 1

Please note that a3=a3 , b3=b3 and c3=c3

Solution:

Let f(a)= a3+b3+c3−3abc be a function in a.

Now, putting a = -(b+c), we get

f( -(b+c)) = b3+c3 - (b+c)3 + 3bc(b+c) = (b+c)3 - (b+c)3 = 0

Hence, by factor theorem, (a+b+c) is a factor.

Note that the expression a3+b3+c3−3abc is homogenous wrt a, b, and c hence no linear factors exist, as the degree of the expression is 3 and we have a factor of degree 1, the other factor must be of degree 2.

Hence we have the other factor = (a2 + b2 + c2) + k (ab + bc + ca) ; where k is any integer (since net coefficients are integers).

Now ((a2 + b2 + c2) + k (ab + bc + ca) ) (a+b+c) = a3+b3+c3−3abc.

The value of can be easily found out to be -1 (even by simply multiplying and comparing); hence the other factor, (a2 + b2 + c2 - ab - bc - ca) .

Thus a3+b3+c3−3abc = (a2 + b2 + c2 - ab - bc - ca) (a+b+c).

Note : The factor theorem usually helps in such questions only for finding homogenous factors or a single linear factor. It is partly guess work on what values should be put in f(a).

• 3
(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
• -3

• -2
a3 + b3​ + c​3 - 3abc
=[ a3 + b3 ]+c3 - 3abc
=[ (a+b)3 - 3ab(a+b) ] + c3 -3abc                    [ Put (a+b) = x ]
=x3​ - 3abx + c- 3abc
​=x3 + c- 3abx - 3abc
=(x + c)(x+ c- cx) - 3ab(x + c)
=(x + c) [ (x+ c- cx - 3ab) ]                        [ Replace x = a+b ]
=(a+b+c) [ (a+b) + c2- c(a+b) - 3ab ]
=(a+b+c) [ a2+ b2+ 2ab + c- ca- cb - 3ab ]
=(a+b+c) [a+ b+ c2- ab - bc - ca ]
• 12
(a2 + b2+c2 -ab -bc -ac) (a+b+c)
• 2

• -4
(a+b+c)(ab-bc-ca-c2+b2+a2
• -2
(a+b)3
• -4
its -1

• 0
(a+b)3

• 0
I think using this formula you should find that in easy way • -5
Yes x = x to the power 10000
• 0
(a+b+c)*(a​2+b2+c​2-ab-bc-ca)
​IS CORRECT
• 2
Ans
• -1
This pic • 5
(a+b+c)(a2+b2+c2-ab-bc-ca)
• 4
(a+b+c)(a2+b2+c2-a.b-b.c-c.a)
• 0
this question is not appropriate ok just check it out once more carefully better for your stuff

• 0
(a + b + c)(a2 + b2 + c2 - ab - bc - ca )
• 2
(a3+b3+c3) +3a2bc+3ab2c+3abc2
• 0 • 0 • -1
(a+b+c)3
• -1
the resultt is 0 when a=b=c=0
• 0
Following ans. • 0
Question 80 • -1
(a+b+c-ab-bc-ca)
• 2 • 0
1/2(a+b+c)(a-b^2+b-c^2+c-a^2)
• 2
NAILED IT

• 2
esdsdg
• 1
(a+b+c)(a2+b2+c2-ab-bc-ca
• 1
A+b+c ka hole square
• 0
(a+b+c)(a2+b2+c2-ab-bc-ac)
• 0
a^3 -b^3

• 0
a3 - b3
​​​​​​​
• 0
Lamba land dede
• 0
hey the answer is ,........ (a+b+c)(a2+b2+c2-ab-bc-ca) hope it hlpz u ....??
• 2
Fdjxg
• 0
(a+b+c)^3
• 0
Srry bro for before answer of mine its original answer is a+b+c=0
• 1
Ans: a^3+b^3+c^3- 3 abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
• 2
abcd
• 2
I hope we study the equation in class 10th. Here is the answer if this was the question • 1
3(a+b+c-abc)
• 0
=3(a+b+c-abc)
• 0
type the same on google by copy paste.
• 1
(a+b+c)^3 - (ab+bc+ca)
• 0 • 1
3w+3b+3c-3abc=o
• 1 • 0
jhfzuj
• 0
Zero
• 0
please give me solution of Q1(i) • 0
Mere questions ka javab do koi toh ylll
• 0 • 0
(A+B+C)(A^2+B^2+B^2-AB-BC-CA)
• 0
(a+b+c) (a^2+b^2+c^2+ab+bc+ca)
• 0
[a+b+c][a2+b2+c2-ab-bc-ca]
• 0
Complete the question
• 0
true or false mass is a vector quantity
• 0 • 0
A2+b2+c2
• 0
????? ?????? ???? ??? ???? 2617465
• 0
B3+a
• 0
(-2,2)(0,-4)
• 0 