show that $a^{3}+b^{3}+c^{3}-3 a b c=1$

Asked Jun 14 '20 at 15:02

Active Jun 14 '20 at 15:02

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$$ \begin{aligned} \text { If } \quad a &=1+\frac{x^{3}}{3 !}+\frac{x^{6}}{6 !}+\ldots \\ & b=x+\frac{x^{4}}{4 !}+\frac{x^{7}}{7 !}+\ldots \\ & c=\frac{x^{2}}{2 !}+\frac{x^{5}}{5 !}+\frac{x^{8}}{8 !}+\ldots \end{aligned} $$ then show that $a^{3}+b^{3}+c^{3}-3 a b c=1$

My Aprroach:

$$\begin{array}{l} a=1+\frac{x^{3}}{3 !}+\frac{x^{2}}{6 !} \\ b=x+\frac{x^{4}}{4 !}+\frac{x^{7}}{7 !} \\ \begin{array}{ll} c=\frac{x^{2}}{2 !}+\frac{x^{5}}{5 !}+\frac{x^{8}}{8 !} \\ \text { Now, } & a+b+c=e^{x} \\ & \frac{1}{2}(a+b+c)\left\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right. \\ & \frac{1}{2} e^{x}({........}) \end{array} \end{array}$$

Further I am not getting no clue!

asked Jun 14 '20 at 15:02

1

Does this answer your question? How to prove $a^3 + b^3 +c^3 - 3abc =1 $, If a,b,c are infinite series as given in the description box below? – found with Approach0 – Martin R Jun 14 '20 at 15:04
$$(a+b+c)(a+\omega b+\omega^2 c)(a+\omega^2 b+\omega c)=a^3+b^3+c^3-3abc$$ where $\omega=\exp(2\pi i/3)$. – Angina Seng Jun 14 '20 at 15:09
yes! thanks a lot – Jun 14 '20 at 15:18

show that $a^{3}+b^{3}+c^{3}-3 a b c=1$

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