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I believe it is (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) Because (a + b + c)(a^2 + b^2 + c^2 + x) a^3 + ab^2 + ac^2 + ax ba^2 + b^3 + c^2b + bx
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Art. 23. ALGEBRA may be defined, a general method of investigating the relations of quantities, by letters, and other symbols. This, it must be acknowledged, is an imperfect...
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If you put a = -b-c, you can see that the expression is zero. (-b-c)^3+b^3+c^3-3(-b-c)bc =-b^3-3b^2c-3bc^2-c^3 +b^3+c^3+3b^2c+3bc^2 =0 So a+b+c is a factor.
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Hence by factor theorem, (a+b+c) is a factor of a^3 + b^3 + c^3 - 3abc 3) Thus a^3 + b^3 + c^3 - 3abc â (a+b+c)*{p(a^2 + b^2 + c^2) + q(ab + bc + ca)}
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a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) applied in many ... factor of a^3 + b^3 + c^3 - 3abc gives . a^3 + b^3 + c^3 - 3abc = r^3 + 2s^3 + 4t^3 - 3[r][s*2^(1 ...
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factor a 3 b 3 c 3 3abc Top 10 from Entireweb
| Algebraic Notation and Coefficients |
| Art. 23. ALGEBRA may be defined, a general method of investigating the relations of quantities, by letters, and other symbols. This, it must be acknowledged, is an imperfect... |
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factor a 3 b 3 c 3 3abc Top 10 from Bing
| Can you factor this? a^3 + b^3 + c ^3 - 3abc? - Yahoo! Answers |
| I believe it is (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) Because (a + b + c)(a^2 + b^2 + c^2 + x) a^3 + ab^2 + ac^2 + ax ba^2 + b^3 + c^2b + bx |
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| Use factor theorem to factorise a^3 + b^3 + c^3 - 3abc? - Yahoo ... |
| If you put a = -b-c, you can see that the expression is zero. (-b-c)^3+b^3+c^3-3(-b-c)bc =-b^3-3b^2c-3bc^2-c^3 +b^3+c^3+3b^2c+3bc^2 =0 So a+b+c is a factor. |
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| Simplify a^3 + b^3 + c^3? - Yahoo!7 Answers |
| Hence by factor theorem, (a+b+c) is a factor of a^3 + b^3 + c^3 - 3abc 3) Thus a^3 + b^3 + c^3 - 3abc â (a+b+c)*{p(a^2 + b^2 + c^2) + q(ab + bc + ca)} |
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| Factorization of a^3 + b^3 + c^3 - 3abc - sci.math | Google ... |
| a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) applied in many ... factor of a^3 + b^3 + c^3 - 3abc gives . a^3 + b^3 + c^3 - 3abc = r^3 + 2s^3 + 4t^3 - 3[r][s*2^(1 ... |
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| Factorization of a^3 + b^3 + c^3 - 3abc |
| I've seen the identity a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) applied in many ways (solving cubic equations, obtaining the cubic arithmetic and ... |
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| Yahoo! Answers - How do i factor this? a^3 + b^3 - ab (a+b)? |
| Then, factor out a + b... (a + b)(aò - ab + bò - ab) = (a + b)(aò - 2ab + bò) ... = a^3 + b^3 + c^3 - 3abc (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) = a^3 + b^2a + c^2a ... |
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| Can you factor this? a^3 - b^3 - c ^3 - 3abc? [Jun 26, 2009] Best Answer: I believe it is ( a - b - c)(a^2 - b^2 - c^2 - ab - bc - ac) Because ( a - b - c)(a^2 - b^2 - c^2 - x) a^ 3 ... |
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| (For example, can you factor $a^3 + b^3 + c^3 - 3abc?$ Think about what the Factor Theorem from precalculus tells you for $x^3 + b^3 + c^3 - 3xbc$ when $x = -b - c.$) See also the ... |
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| Fun with maths: integers x = a3+ b3+ c3-3abc for some integers a ... |
| a^3+ b^3+ c^3-3abc = (a+bw+cw^2)(a+bw^2+cw) where w = cube root of -1 let ... factor by^2-b^2y-ay^2+a^2y+ab^2-ba^2; solve x = a/(b+c) = b/(c+a) = c/(a+b) |
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| Plz solve: If a+b+c=1,a^2+b^2+c^2=2,a^3+b^3+c^3=3, find the value ... |
| a^3 +b^3 +c^3 -- 3abc = (a+b+c) {a^2+b^2+c^2 --(ab+bc+ca)}--->(2) (a^2 + b^2 + c^2)^2 = ... How do you factor "a^3-b^6"? what is (a^(1/3)-b^(1/3)(a^2/3)+[ab]^(1/3)+b^(2/3 ... |
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factor a 3 b 3 c 3 3abc Top 10 from SearchHippo
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