Complex

Calculus Level 4

A = 1 + x 3 3 ! + x 6 6 ! + x 9 9 ! + ⋯ B = x + x 4 4 ! + x 7 7 ! + x 10 10 ! + ⋯ C = x 2 2 ! + x 5 5 ! + x 8 8 ! + x 11 11 ! + ⋯ \begin{aligned} A & =1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots \\ B & =x+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^{10}}{10!}+\cdots \\ C& =\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\cdots \end{aligned}

Given the above, what is A 3 + B 3 + C 3 − 3 A B C + 3 A^3+B^3+C^3-3ABC+3 ?


The answer is 4.

Digvijay Singh by Digvijay Singh , 21, India

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1 solution

Digvijay Singh
Jan 21, 2018

Let 1 , ω 1, \omega and ω 2 {\omega}^2 be the cube roots of unity.

Then, it can be noted that

  • A + B + C = e x \large A+B+C=e^x

  • A + ω B + ω 2 C = e ω x \large A+\omega B+{\omega}^2 C=e^{\omega x}

  • A + ω 2 B + ω C = e ω 2 x \large A+{\omega}^2 B+\omega C=e^{{\omega}^2 x}

A 3 + B 3 + C 3 − 3 A B C \large A^3+B^3+C^3-3ABC

= ( A + B + C ) ( A 2 + B 2 + C 2 − A B − B C − A C ) \large =(A+B+C)(A^2+B^2+C^2-AB-BC-AC)

= ( A + B + C ) ( A + ω B + ω 2 C ) ( A + ω 2 B + ω C ) \large = (A+B+C)(A+\omega B+{\omega}^2 C)(A+{\omega}^2 B+\omega C)

= e x ⋅ e ω x ⋅ e ω 2 x \large = e^x\cdot e^{\omega x}\cdot e^{{\omega}^2 x}

= e ( 1 + ω + ω 2 ) x \large = e^{(1+\omega+{\omega}^2)x}

= e 0 = 1 \large=e^0= 1

Which makes the answer 4 \boxed{4}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

can you explain how u went from line

a^3+b^3+^3-3abc to line with w w^2 (right below)

thanks

D S - 3 years, 4 months ago

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i factorised A 3 + B 3 + C 3 − 3 A B C A^3+B^3+C^3-3ABC .

Digvijay Singh - 3 years, 4 months ago

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