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the quantity 4+radical 8 times the quantity 4 minus 4 radical 8.
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\[{{(4+\sqrt{8})} \times {(4-{4}\sqrt{8})}}\]
is this your question?
yes it is
plz see the second term once
take to second 4 away ,just leave one 4
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\[{4^2 - \sqrt{8^2}}= 16 - 8 = 8\]
(4 + sqrt8)*(4-sqrt8) = 4^2 -(sqrt8)^2 = 16 - 8 = 8
thanks :)
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