may i please have help solving with the following 2 equations:
$\displaystyle
3b^3 \sqrt{20 a^3 b^2 c^5}
$
and
$\displaystyle \sqrt{3}(4+\sqrt{3}- \sqrt{15})$
What you have are NOT equations. You only have an equation when you have an equals sign.
I think you're asking to simplify these expressions.
For the first one:
$\displaystyle 3b^3\sqrt{20 a^3 b^2 c^5} = 3b^3\times\sqrt{20}\times\sqrt{a^3}\times\sqrt{b^2 }\times\sqrt{c^5}$
$\displaystyle = 3b^3\times\sqrt{4\times 5}\times\sqrt{a^2\times a}\times b \times \sqrt{c^4\times c}$
$\displaystyle = 3b^4 \times \sqrt{4}\times\sqrt{5}\times\sqrt{a^2}\times\sqrt{ a}\times \sqrt{c^4} \times \sqrt{c}$
$\displaystyle = 3b^4 \times 2 \times \sqrt{5} \times a \times \sqrt{a} \times c^2 \times \sqrt{c}$
$\displaystyle = 6ab^4c^2 \times \sqrt{5} \times \sqrt{a}\times \sqrt{c}$
$\displaystyle = 6ab^4c^2\sqrt{5ac}$
The second one looks already simplified. You can't break up a surd over addition and subtraction (only multiplication and division) and there aren't any other common factors besides the 3.
I think the only other ways it could be written are
$\displaystyle \sqrt{3}\sqrt{4 + \sqrt{3} - \sqrt{15}}$
or
$\displaystyle \sqrt{12 + 3\sqrt{3} - 3\sqrt{15}}$.
The OP's 2nd question is different than what has been discussed:
$\displaystyle \sqrt{3}(4+\sqrt{3}- \sqrt{15})$
Assuming that this is the actual question, to solve this you distribute the square root of 3:
$\displaystyle 4\sqrt{3} + \sqrt{3}\sqrt{3} - \sqrt{3}\sqrt{15}$
$\displaystyle 4\sqrt{3} + 3 - \sqrt{45}$
$\displaystyle 4\sqrt{3} + 3 - 3\sqrt{5}$
$\displaystyle 3 + 4\sqrt{3} - 3\sqrt{5}$
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