Calvin learns Pythagorean triplets!

Algebra Level 2

$\Large{(3^3+4^3+5^3)^{\frac{4}{3}}}$

Once Calvin learnt the famous Pythagoras Theorem and he came to know about the triplet $(3,4,5)$ . So instead of squaring them , he cubed them to find an expression below. So what was the value of the above expression that Calvin had found?

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The answer is 1296.

4 solutions

Rohit Ner
Jun 18, 2015

$\begin{aligned}(3^3+4^3+5^3)^{\frac{4}{3}} & =(3.3^2+4.4^2+5.5^2)^{\frac{4}{3}} \\ & = (4^2+3.5^2+5.5^2)^{\frac{4}{3}} \\ & =(4^2+8.5^2)^{\frac{4}{3}} \\ & =(16+200)^{\frac{4}{3}} \\ & =216^{\frac{4}{3}} \\ & \Huge{\color{#3D99F6}{=\boxed{1296}}}\end{aligned}$

Moderator note:

Yes correct. Isn't the equation $3^3 + 4^3 + 5^3= 6^3$ beautiful?

Sorry, could you explain me please what you did on step two?

Esteban Vasquez Giraldo - 5 years, 12 months ago

By Pythagoras Theorem, $3^2 + 4^2 + 5^2$ , so $3\cdot 3^2 + 4 \cdot 4^2 = 3\cdot 3^2 + 3 \cdot 4^2 + 4^2 = 3(3^2+ 4^2) + 4^2 = 3\cdot 5^2 + 4^2$ .

Pi Han Goh - 5 years, 12 months ago

Thank you!

Esteban Vasquez Giraldo - 5 years, 12 months ago

Sorry, in some reason i still can't understand. Can you explain it with more detail ?

Daniel Sugihantoro - 5 years, 9 months ago

@Daniel Sugihantoro – Which part?

Pi Han Goh - 5 years, 9 months ago

@Pi Han Goh – 3 x 3^2 + 4 x 4^2 = 3 x 4^2 + 3 x 4^2 + 4^2

how could it be ?

Daniel Sugihantoro - 5 years, 8 months ago

In response to Challenge Master : yeah it is ! Are there more such quadruplets ?

raman rai - 5 years, 11 months ago

thanks bro

గిరీష్ కండలవీరుడు - 5 years, 11 months ago

(3^3+4^3+5^3)^4/3 =(27+64+125)^4/3 =(216)^4/3 =725594112

Habib Musabbir - 5 years, 11 months ago

$\begin{aligned}(3^3+4^3+5^3)^{\frac{4}{3}} & =(3.3^2+4.4^2+5.5^2)^{\frac{4}{3}} \\ & = (4^2+3.5^2+5.5^2)^{\frac{4}{3}} \\ & =(4^2+8.5^2)^{\frac{4}{3}} \\ & =(8.2+8.25)^{\frac{4}{3}} \\ & =(8.27)^{\frac{4}{3}} \\ & =(2.3)^{4} \\ & = 6^{4} \\ & \Huge{\color{#3D99F6}{=\boxed{1296}}}\end{aligned}$

Andi Setiawan - 5 years, 9 months ago

Rama Devi
Jun 22, 2015

It can be noticed that $3^3+4^3+5^3=6^3$ . Therefore the answer is $6^4$ , which is $1296$

Muhammad Humaiz Anjum
Sep 27, 2015

$\Large{(3^3+4^3+5^3)^{\frac{4}{3}}}\\= \Large{(27+64+125)^{\frac{4}{3}}}\\=\Large{(216)^{\frac{4}{3}}}=\Large{(6^{3})^{\frac{4}{3}}}=\Large(6)^{4} =\Large{1296}$

Jesse Nieminen
Jun 23, 2015

(3^3 + 4^3 + 5^3)^4/3

= ((3)^3 + (3+1)^3 + (3+2)^3)^4/3

= ((3^3) + (3^3 + 3 * 3^2 + 3 * 3 + 1) + (3^3 + 6 * 3^2 + 12 * 3 + 8))^4/3

= (3 * 3^3 + 9 * 3^2 + 15 * 3 + 9)^4/3

= (9 * 3^2 + 9 * 3^2 + 5 * 3^2 + 3^2)^4/3

= (24 * 3^2)^4/3

= (2^3 * 3^3)^4/3

= (6^3)^4/3

= 6^4

= 1296

Calvin learns Pythagorean triplets!

Click here for full set .

The answer is 1296.

4 solutions

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