Question

Write down the he square of ( 7x cube + 1 upon 7y cube )

Answer 1

Answer:

$(7 {x}^{3} + \frac{1}{7 {y}^{3} } ) {}^{2} = 49 {x}^{6} + 2 \times 7 {x}^{3} \times \frac{1}{7 {y}^{3} } + \frac{1}{49 {y}^{6} }$

$= 49 {x}^{6} + 2 \frac{ {x}^{3} }{ {y}^{3} } + \frac{1}{49 {y}^{6} }$