# Step-by-step Solution

## Solve (2a*b*c-a*b*c^0.5)^2

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### Videos

$4a^2b^2c^2-4a^2b^2\sqrt{c^{3}}+a^2b^2c$

## Step-by-step explanation

Problem to solve:

$\left(2abc-\sqrt{abc}\right)^2$
1

A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: $(a-b)^2=a^2-2ab+b^2$

• Square of the first term: $\left(2abc\right)^2 = 4a^2b^2c^2$
• Double product of the first by the second: $2\left(2abc\right)\left(-ab\sqrt{c}\right) = -4abcab\sqrt{c}$
• Square of the second term: $\left(-ab\sqrt{c}\right)^2 = a^2b^2c$

$4a^2b^2c^2-4abcab\sqrt{c}+a^2b^2c$
2

When multiplying exponents with same base you can add the exponents

$4a^2b^2c^2-4c\sqrt{c}a^2b^2+a^2b^2c$

$4a^2b^2c^2-4a^2b^2\sqrt{c^{3}}+a^2b^2c$
$\left(2abc-\sqrt{abc}\right)^2$