FACTORIZE : A³

     
Find one factor of the size a^k+m, where a^k divides the monomial with the highest power nguồn a^3 and m divides the constant factor b^3+c^3. One such factor is a+b+c. Factor the polynomial by dividing it by this factor.

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I suggests that you use (a+b)^3=a^3+b^3+3ab(a+b)Rightarrow a^3+b^3=(a+b)^3-3ab(a+b) instead, you will need lớn use it twice lượt thích this: a^3+b^3+c^3-3abc =(a+b)^3+c^3-3ab(a+b)-3abc =(a+b+c)^3-(3c(a+b)^2+3(a+b)c^2)-3ab(a+b+c) ...
How khổng lồ prove a^3 + b^3 +c^3 - 3abc =1 , If a,b,c are infinite series as given in the description box below?
If all three expressions are considered as functions of x in Bbb R (or Bbb C, if you prefer): a(x) = 1+fracx^33!+fracx^66! + ldots ,\ b(x)=x+fracx^44!+fracx^77!+ ldots ,\ c(x)=fracx^22!+fracx^55!+fracx^88!+ ldots ...
If you require a, b, c distinct, no. The term a+b+c can't be 0 as they are all the same sign & the other term is a sum of squares which can only be 0 if a=b=c.

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What axioms are you using? How was p closed under multiplication proven? I think your answers are probably correct if those indeed are the axioms & properties actually available lớn you (they ...
https://math.stackexchange.com/questions/2252741/if-one-root-of-the-equation-ax2bxc-0-be-the-square-of-the-other-then-which
Vieta's formula says: alpha+alpha^2=-dfrac ba,alphacdotalpha^2=dfrac ca left(-dfrac ba ight)^3=(alpha+alpha^2)^3=alpha^3+(alpha^3)^2+3alpha^3left(-dfrac ba ight) Replace alpha^3 ...
By the AM-GM inequality : fraca^3+b^3+c^33 ge sqrt<3>a^3b^3c^3 = abc The strict inequality holds unless ,a=b=c,.

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Find one factor of the form a^k+m, where a^k divides the monomial with the highest power a^3 and m divides the constant factor b^3+c^3. One such factor is a+b+c. Factor the polynomial by dividing it by this factor.
left< eginarray l l 2 & 3 \ 5 & 4 endarray ight> left< eginarray l l l 2 & 0 và 3 \ -1 & 1 và 5 endarray ight>
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