Question

(6√27-4√100)*(6√27+4√100)​

Answer 1

$\displaystyle\bf\\\boxed{(a-b)\cdot(a+b)=a^{2} -b^{2} }\\\\\\(6\sqrt{27} -4\sqrt{100})\cdot(6\sqrt{27}+4\sqrt{100})=\\\\\\=(6\sqrt{27})^{2} -(4\sqrt{100})^{2} =36\cdot 27-16\cdot 100=\\\\\\=972-1600=-628$

Answer 2

To simplify the expression $(6\sqrt{27}-4\sqrt{100}) \times (6\sqrt{27}+4\sqrt{100})$, we can use the difference of squares formula, which states that $a^2 - b^2 = (a + b)(a - b)$. Let's apply this to the given expression step by step.

First, let's simplify each term:

$6\sqrt{27} = 6 \times 3\sqrt{3} = 18\sqrt{3}$

$4\sqrt{100} = 4 \times 10 = 40$

So the expression becomes:

$(18\sqrt{3} - 40) \times (18\sqrt{3} + 40)$

Now, we can apply the difference of squares formula:

$(18\sqrt{3} - 40) \times (18\sqrt{3} + 40) = (18\sqrt{3})^2 - 40^2$

$= (18\sqrt{3})^2 - 1600$

$= 324 \times 3 - 1600$

$= 972 - 1600$

$= -628$

So the simplified expression is $-628$.