what is (a+b+c)^3?
please answer fast anybody make sure that you do your best to get the exact answer
Answers
Mentally examine the expansion of (x+y+z)3(x+y+z)3 and realize that each term of the expansion must be of degree three and that because x+y+zx+y+z is cyclic all possible such terms must appear. Those types of terms can be represented by x3,x2yx3,x2y and xyzxyz. If x3x3 appears, so must y3y3 and z3z3. If x2yx2y appears, then so must x2z,y2x,y2z,z2xx2z,y2x,y2z,z2x and z2yz2y. To top it off, xyzxyz is the final term to be included. Therefore
(x+y+z)3=x3+y3+z3+m(x2y+x2z+y2x+y2z+z2x+z2y)+nxyz.(x+y+z)3=x3+y3+z3+m(x2y+x2z+y2x+y2z+z2x+z2y)+nxyz.
We need only deduce the values of mm and nn. To build an xyzxyz term, there are 3 ways to select the xx and then 2 ways for the yy so n=6n=6.
Replacing x,yx,y and zz with 1, the left side becomes 2727 and the right side becomes 9+6m9+6m. Therefore m=3m=3. Now we have it
(x+y+z)3=x3+y3+z3+3(x2y+x2z+y2x+y2z+z2x+z2y)+6xyz.(x+y+z)3=x3+y3+z3+3(x2y+x2z+y2x+y2z+z2x+z2y)+6xyz.
Remember this! Now make the replacements x=a,y=bx=a,y=b and z=−cz=−c.
It seems as though we’ve done a lot of work there. Not true. When expanding such a thing for fun or fortune I do not write 1137 words, I just write the answer.
plz follow me guys plz plz
Hey mate❤
find here
(a+b+c)³=..❓❓
so,
we know
(a+b+c)³= a³+b³+c³+3(a+b)(b+c)(c+a)
proof :--
(a+b+c)³= a³+b³+c³+3(a+b)(b+c)(c+a)
it can we written as
(a+b+c)³= a³+b³+c³+3(a+b)(b+c)(c+a)
=>(a+b+c)³-a³-b³-c³ = 3(a+b)(b+c)(c+a). ...........(1)
consider LHS ....
so,
(a+b+c)³-a³-b³-c³
= [a³+b³+c³+3ab(a+b)+3bc(b+c)+3ca(c+a)+6abc-a³-b³-c³
=3ab(a+b)+3bc(b+c)+3ca(c+a)+6abc
=3 [ab(a+b)+bc(b+c)+ca(c+a)+2abc]
=3[ab(a+b)+b²c+bc²+c²a+ca²+2abc]
=3[ab(a+b)+(abc+b²c)+(abc+ca²)+(c²a+bc²)]
=3[ab(a+b)+bc(a+b)+ca(b+a)+c²(a+b)]
= 3(a+b)(ab+BC+ca+c²)
=3(a+b)[a(b+c)+c(b+c)]
=3(a+b)(a+c)(b+c)
Which is equals RSH of equation (1)
hopes its help's u
➡Follow me
@abhi.✌✌