How can I write (x³ - (3/8))(x³ + (16/17)) in the standard form?

Answer:
$\left(x^3 - \frac{3}{8}\right)\left(x^3 + \frac{16}{17}\right)$

$\Rightarrow$ $\left( \left(x^{3 }\right) \left(x^3\right) + \left(\frac{16}{17}{x}^3- \frac{3}{8}{x}^3 \right) -\left(\frac{3}{8}\right)\left(\frac{16}{17}\right) \right)$

Taking L.C.M of middle term and get As:


$\Rightarrow$ $\left(x^6 + \frac{128 x^3 - 51 x^3}{136} -\frac{48}{136} \right)$   

Taking L.C.M. and get

$\Rightarrow$ $\left(\frac{136x^6 + 77 x^3 - 48}{136} \right)$

Equate equal to Zero and we get

$\Rightarrow$ $\left(\frac{136x^6 + 77 x^3 - 48}{136} \right)$ = 0

$\Rightarrow$ 136 x⁶ + 77 x³ - 48   = 0                                      ( Ans )