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Number Theory Level 2

288 3 × 432 3 × 648 3 = ? \Large \color{#624F41}{\sqrt[3]{288} \times \sqrt[3]{432} \times \sqrt[3]{648}=\ ?}

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Some Irrational Number. 288 None of the given choices. 432 648

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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3 solutions

Nihar Mahajan
May 14, 2015

288 3 × 432 3 × 648 3 = 2 5 3 2 3 × 2 4 3 3 3 × 2 3 3 4 3 = 2 12 3 9 3 = 2 4 3 3 = 432 \Large{\sqrt[3]{288} \times \sqrt[3]{432} \times \sqrt[3]{648} \\= \sqrt[3]{2^53^2} \times \sqrt[3]{2^43^3} \times \sqrt[3]{2^33^4} \\ =\sqrt[3]{2^{12}3^9} \\ =2^43^3 \\ =\large\boxed{432}}

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Moderator note:

Wonderful factorization of numbers. Bonus question: Without using a calculator, evaluate the expression below.

( 288 3 + 432 3 + 648 3 ) × [ ( 288 3 − 432 3 ) 2 + ( 432 3 − 648 3 ) 2 + ( 648 3 − 288 3 ) 2 ] \left ( \sqrt[3]{288} + \sqrt[3]{432} + \sqrt[3]{648} \right ) \\ \times \\ \left [ \left( \sqrt[3]{288} - \sqrt[3]{432} \right)^2 + \left( \sqrt[3]{432} - \sqrt[3]{648} \right)^2+ \left( \sqrt[3]{648} - \sqrt[3]{288}\right)^2 \right ]

[Response to Challenge Master Note]

Let a = 288 3 , b = 432 3 a=\sqrt[3]{288}~,~b=\sqrt[3]{432} and c = 648 3 c=\sqrt[3]{648} . The given expression (call it E \mathcal{E} ) to be evaluated is,

E = ( a + b + c ) [ ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 ] = 2 ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) = 2 ( a 3 + b 3 + c 3 − 3 a b c ) = 2 ( 288 + 432 + 648 − 3 × 432 ) = 144 \begin{aligned}\mathcal{E}&=(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\\&=2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\&=2(a^3+b^3+c^3-3abc)\\&=2(288+432+648-3\times 432)=\boxed{144}\end{aligned}

Prasun Biswas - 6 years ago

Response to Nihar Mahajan: Check this one. The equation is in form of X.3X/2.9X/4 whole cube rooted. Now simple multiplication gives us 27.X^3/8 cube rooted. removing cube root. gives us 3X/2. which is the middle number 432

Kaleem Ullah - 6 years ago
Mehul Arora
May 14, 2015

We know that,

a 3 ∗ b 3 ∗ c 3 = a b c 3 \sqrt[3]{a}*\sqrt[3]{b}*\sqrt[3]{c}=\sqrt[3]{abc}

Now, Note that:-

288 = 2 ∗ 12 ∗ 12 288=2*12*12

432 = 6 ∗ 6 ∗ 12 432=6*6*12

648 = 2 ∗ 2 ∗ 6 ∗ 27 648=2*2*6*27

Therefore,

288 ∗ 432 ∗ 648 3 = 2 ∗ 2 ∗ 2 ∗ 12 ∗ 12 ∗ 12 ∗ 6 ∗ 6 ∗ 6 ∗ 3 ∗ 3 ∗ 3 3 \sqrt[3]{288*432*648}= \sqrt[3]{2*2*2*12*12*12*6*6*6*3*3*3}

2 3 ∗ 3 3 ∗ 6 3 ∗ 12 3 3 = ( 432 3 3 = 432 \sqrt[3]{{2}^{3}*{3}^{3}*{6}^{3}*{12}^{3}}= (\sqrt[3]{{432}^{3}}= 432

Hence, The answer is 432 \boxed{432}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Nice way to factorizing the terms "half way". Good work! Bonus question: Because it's only true that A × B = A B \sqrt A \times \sqrt B = \sqrt {AB} when A , B ≥ 0 A,B \geq 0 ; is it also true that A 3 × B 3 = A B 3 \sqrt[3]{ A} \times \sqrt[3]{ B} = \sqrt[3]{AB} only when A , B ≥ 0 A,B \geq 0 ?

Mrh Riyad
May 22, 2015

288 3 × 432 3 × 648 3 \Large\color{#D61F06}{\sqrt[3]{288}\times\sqrt[3]{432}\times\sqrt[3]{648}} = 288 × 432 × 648 3 \Large\color{#D61F06}{=\sqrt[3]{288\times432\times648}} = 144 × 2 × 432 × 216 × 3 3 \Large\color{#D61F06}{=\sqrt[3]{144\times2\times432\times216\times3}} = 144 × 3 × 432 × 216 × 2 3 \Large\color{#D61F06}{=\sqrt[3]{144\times3\times432\times216\times2}} = 432 × 432 × 432 3 \Large\color{#D61F06}{=\sqrt[3]{432\times432\times432}} = ( 3 432 ) 3 \Large\color{#D61F06}{=\sqrt[3](432)^3} = 432 \Large\color{#D61F06}{=\boxed{432}}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

I wasn't expecting this. Nicely done!

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