Question

proove it. 27 ( a + b + c )³ - (2a + b )³ - ( 2b + c )³ - (2c + a )³ = 3 ( a + 2b +3c ) ( b + 2c + 3a ) ( c + 2a + 3b )​

Answer 1

Answer:

>>>> Mentally examine the expansion of (x+y+z)3 and realize that each term of the expansion must be of degree three and that because x+y+z is cyclic all possible such terms must appear. Those types of terms can be represented by x3,x2y and xyz . If x3 appears, so must y3 and z3 . If x2y appears, then so must x2z,y2x,y2z,z2x and z2y . To top it off, xyz is the final term to be included. Therefore

>>>(x+y+z)3=x3+y3+z3+m(x2y+x2z+y2x+y2z+z2x+z2y)+nxyz.

We need only deduce the values of m and n . To build an xyz term, there are 3 ways to select the x and then 2 ways for the y so n=6 .

Replacing x,y and z with 1, the left side becomes 27 and the right side becomes 9+6m . Therefore m=3 . Now we have it

(x+y+z)3=x3+y3+z3+3(x2y+x2z+y2x+y2z+z2x+z2y)+6xyz.

˙·٠•●♥ REMEMBER THIS! NOW MAKE THE REPLACEMENTS X=A,Y=B AND Z=−C .

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