Question 767162
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64s<sup>6</sup>+t<sup>6</sup>

That's the sum of two cubes:

64s<sup>6</sup> is equal to (4s<sup>2</sup>)<sup>3</sup>

t<sup>6</sup> is equal to (t<sup>2</sup>)<sup>3</sup>

The sum of two cubes A<sup>3</sup>+B<sup>3</sup> factors as (A+B)(A<sup>2</sup>-AB+B<sup>2</sup>)

(A+B)(A<sup>2</sup>-AB+B<sup>2</sup>)

let A = 4s<sup>2</sup>, B = t<sup>2</sup>

(A+B)(A<sup>2</sup>-AB+B<sup>2</sup>) = (4s<sup>2</sup>+t<sup>2</sup>)[(4s<sup>2</sup>)<sup>2</sup>-(4s<sup>2</sup>)(t<sup>2</sup>)+(t<sup>2</sup>)<sup>2</sup>] = 

(4s<sup>2</sup>+t<sup>2</sup>)[16s<sup>4</sup>-4s<sup>2</sup>t<sup>2</sup>+t<sup>4</sup>]  

(4s<sup>2</sup>+t<sup>2</sup>)(16s<sup>4</sup>-4s<sup>2</sup>t<sup>2</sup>+t<sup>4</sup>)

Edwin</pre>