Sanhita M. answered • 01/25/16
Tutor
4.7
(11)
Mathematics and Geology
Lets do it:
a3 + b3 + c3 - 3abc
=(a+b+c)3-3(a+b)2c-3(a+b)c2-3(a+b)ab-3abc ..... since by formula (a+b+c)3=a3 + b3 + c3+3(a+b)2c+3(a+b)c2 +3a2b + 3ab2, to have only a3 + b3 + c3 from (a+b+c)3 the rest, i.e., [3(a+b)2c+3(a+b)c2+3a2b + 3ab2]must be subtracted.Thus, it is pure algebraic adjustment. Then next line will be:
=(a+b+c)3-3(a+b)2c-3(a+b)c2-3(a+b)ab-3abc ..... since by formula (a+b+c)3=a3 + b3 + c3+3(a+b)2c+3(a+b)c2 +3a2b + 3ab2, to have only a3 + b3 + c3 from (a+b+c)3 the rest, i.e., [3(a+b)2c+3(a+b)c2+3a2b + 3ab2]must be subtracted.Thus, it is pure algebraic adjustment. Then next line will be:
= (a+b+c)3-[3(a+b)2c+3(a+b)c2]-[3(a+b)ab+3abc]
=(a+b+c)3-3(a+b)c[(a+b)+c]-3ab[(a+b)+c]
=(a+b+c)3-3(a+b)c(a+b+c)-3ab(a+b+c)
=(a+b+c)[(a+b+c)2-3(a+b)c-3ab]
=(a+b+c)[a22ab+b2+2bc+2ca+c2-3bc-3ca-3ab] ..... expanding the formula
=(a+b+c)(a2+b2+c2+-ab-bc-ca] .... by adjustment...
This is the simplest form in which the given expression, i.e. a3 + b3 + c3 - 3abc can be re-expressed. Simplest form is the factorized form. Only observation and grasp of basic concepts and definitions [in this case of cube of polynomials and definition of factorization] helps to deduce the result. Practice is the only method to achieve speed and perfection.
Thus for:
(a + b + c)3 - a3 - b3 - c3 just expanding the first compact cube will do the most, i.e.,
(a + b + c)3 - a3 - b3 - c3 just expanding the first compact cube will do the most, i.e.,
=a3+b3+c3+3(a+b)ab+3(a+b)2c+3(a+b)c2 - a3 - b3 - c3 .......... cancelling the cubed terms
=3(a+b)[ab+(a+b)c+c2] .... adjusting operators
=3(a+b)(ab+bc+ca+c2)
(a + b + c)333 - a333 - b333 - c333 needs one to follow the rule of polynomial expansion, (a+b+c)n=an+bn+cn + n(a+b)(n-1)c+n(a+b)c(n-1)+n(a+b)(n-1)ab and method of induction. which in present case may end up with a solution like: 333(a+b)[(a+b)332c+c332+ab(a+b)331]
=333(a+b)[(a+b)331{(a+b)c+ab}+c332]
=333(a+b)[(a+b)331(ab+bc+ca)+c332]
Sarah W.
Also, how is it clear that a term like a2 + b2 + c2 - ab - ac - bc can no longer be factored somehow? What idea am I missing?
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01/26/16
Sanhita M.
It is pretty encouraging as i hear from you.
a2 + b2 + c2 - ab - ac - bc is an expression which can be derived from adjustments on results of squaring (a-b-c) which may go like following:
(a-b-c)2
=(a-b)2-2(a-b)c+c2
=a2 + b2 + c2 - 2ab - 2ca + 2bc
Or,
(a-b-c)2
=[a-(b+c)]2
=a2-2a(b+c)+(b+c)2
=a2 + b2 + c2 - 2ab - 2ca + 2bc
If tried with switching the negative sign among coefficients in the expression like, (a+b-c) or (a-b+c), the squared form always result in squared terms with constant 1 and multiples of coefficients with constant 2 where two of multiples will have same sign while the remaining will be other. Clearly no single adjustment can manipulate the result into further simplicity of first degree coefficients. Thus we have [a2 + b2 + c2 - ab - ac - bc] as one of the simplest form of expression.
Moreover, further study of expansion of different degree of polynomials will reveal that while positive coefficients in polynomials always gives positive coefficients in expansion, the negative coefficients gives a rhythm of positive and negative terms. The odd degree always has a negative term at the end of expansion and the even degree ends with positive term.
For further practice,Elementary Algebra By Henry Sinclair Hall & Samuel Ratcliff Knight is a great guide for school level, say 8th grade to 12th. For Degree polynomials Higher Algebra by same authors can be great help.
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01/26/16
Sarah W.
01/26/16