Question

I solved some of it but now I don't know if you continue it correctly ... To be performed: (lesson: Raducali of order "n")

Answer 1

Answer:

$( \sqrt[4]{7} + 2)( \sqrt[4]{7} - 2)( \sqrt[4]{7} + 4) \\ (first \: two \: is \: in \: the \: form \: of \: (x + y)(x - y) = {x}^{2} - {y}^{?} \\ then { \sqrt[4]{7} }^{2} - {2}^{2} \\ =( \sqrt{7} - 4) \\ so \: ( \sqrt{7} - 4)( \sqrt{7} + 4) \\ = { \sqrt{7} }^{2} - {4}^{2} \\ = 7 - 16 \\ = - 9 \\ thank \: you$

Answer 2

Answer:

-9

Step-by-step explanation:

$( \sqrt[4]{7} + 2)( \sqrt[4]{7} - 2)( \sqrt{7} + 4) \\ = ( { \sqrt[4]{7} }^{2} - {{2} }^{2} )( \sqrt{7} + 4) [because (a+b)(a-b) = a² - b²]\\ = ( \sqrt{7} - 4)( \sqrt{7} + 4) \\ = { \sqrt{7} }^{2} - {4}^{2}[because (a+b)(a-b) = a² - b²] \\ = 7 - 16 \\ = - 9$