Question

(^6√27+^4√64)(^6√27-^4√64)​

Answer 1

$(\sqrt[6]{27}+\sqrt[4]{64})(\sqrt[6]{27}-\sqrt[4]{64} )=(\sqrt[6]{3^{3} }+\sqrt[4]{2^{4}\cdot 2^{2} })(\sqrt[6]{3^{3} }-\sqrt[4]{2^{4}\cdot 2^{2}} ) =\\\\=(\sqrt{3} +2\sqrt{2} )(\sqrt{3} -2\sqrt{2} )=(\sqrt{3})^{2}-(2\sqrt{2} )^{2}=3-8=\boxed{-5}$

Answer 2

Ответ:

Пошаговое объяснение:

Решаем по формуле разность квадратов.

Answer 3

To simplify the expression ($6\sqrt{27} + 4\sqrt{64})(6\sqrt{27} - 4\sqrt{64})$, you can use the difference of squares formula, which states that $a^2 - b^2 = (a + b)(a - b)$.

In your case:

$a = 6\sqrt{27}$ $b = 4\sqrt{64}$

First, simplify $a$ and $b$:

$6\sqrt{27} = 6\sqrt{3^3} = 6 \cdot 3\sqrt{3} = 18\sqrt{3}$

$4\sqrt{64} = 4\sqrt{8^2} = 4 \cdot 8 = 32$

Now, you can rewrite the expression:

$(18\sqrt{3} + 32)(18\sqrt{3} - 32)$

This expression is in the form of $(a + b)(a - b)$, where $a = 18\sqrt{3}$ and $b = 32$. You can apply the difference of squares formula:

$(a + b)(a - b) = a^2 - b^2$

So,

$(18\sqrt{3} + 32)(18\sqrt{3} - 32) = (18\sqrt{3})^2 - 32^2$

Now, calculate the squares:

$(18\sqrt{3})^2 = 18^2 \cdot (\sqrt{3})^2 = 324 \cdot 3 = 972$

$32^2 = 32 \cdot 32 = 1024$

Now, subtract:

$972 - 1024 = -52$

So, $(6\sqrt{27} + 4\sqrt{64})(6\sqrt{27} - 4\sqrt{64}) = -52$.