Find 2 positive integers greater than 1 whose product is $6^{6} + 8^{4} + 27^{4}$ $6^6 + 8^4 + 27^4$ ?

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Find 2 positive integers greater than 1 whose product is $6^{6} + 8^{4} + 27^{4}$ $6^6 + 8^4 + 27^4$ ?

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Answer 1 · 2017-07-27T01:21:38

See a solution process below:

Explanation:

$6^{6} = 46, 656$ $6^6 = 46,656$

$8^{4} = 4, 096$ $8^4 = 4,096$

$27^{4} = 531, 441$ $27^4 = 531,441$

Therefore:

$6^{6} + 8^{4} + 27^{4} = 46, 656 + 4096 + 531, 441 = 582, 193$ $6^6 + 8^4 + 27^4 = 46,656 + 4096 + 531,441 = 582,193$

$582, 193 = 577 \times 1009$ $582,193 = 577 xx 1009$

Answer 2 · 2017-07-27T09:21:50

$6^{6} + 8^{4} + 27^{4} = 577 \cdot 1009$ $6^6+8^4+27^4=577*1009$

Explanation:

Calling $a = 3^{6}$ $a = 3^6$ and $b = 2^{6}$ $b = 2^6$ we have

$6^{6} + 8^{4} + 27^{4} = a \cdot b + b^{2} + a^{2} = {(a + b)}^{2} - a \cdot b$ $6^6+8^4+27^4=a*b+b^2+a^2 = (a+b)^2-a*b$ or using the identity

$u^{2} - v^{2} = (u - v) (u + v)$ $u^2-v^2=(u-v)(u+v)$

${(2^{6} + 3^{6})}^{2} - {(2^{3} \cdot 3^{3})}^{2} = (2^{6} + 3^{6} - 2^{3} \cdot 3^{3}) (2^{6} + 3^{6} + 2^{3} \cdot 3^{3})$ $(2^6+3^6)^2-(2^3*3^3)^2 = (2^6+3^6-2^3*3^3)(2^6+3^6+2^3*3^3)$ or

$6^{6} + 8^{4} + 27^{4} = 577 \cdot 1009$ $6^6+8^4+27^4=577*1009$

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Find 2 positive integers greater than 1 whose product is $6^{6} + 8^{4} + 27^{4}$ $6^6 + 8^4 + 27^4$ ?

2 Answers

Explanation:

Explanation:

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Find 2 positive integers greater than 1 whose product is 6^6 + 8^4 + 27^466+84+2746^6 + 8^4 + 27^4?

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Find 2 positive integers greater than 1 whose product is $6^{6} + 8^{4} + 27^{4}$ $6^6 + 8^4 + 27^4$ ?