Question

2x^2yz-[3x^2-{2y^2-(x^2yz-y^2+z^2)}] solve this question​

Answer 1

Answer:

x^2yz-3x^2+3y^2-z^2

Answer 2

The answer is $x^2yz -3x^2 + 3y^2 - z^2$.

Given

$2x^2yz-[3x^2-({2y^2-(x^2yz-y^2+z^2))]$

To Find

Simplified expression

Solution

Here we have,

$2x^2yz-[3x^2-({2y^2-(x^2yz-y^2+z^2))]$

Since there are no common terms in the round bracket, we will remove them. Due to the minus sign, all the signs are going to change. Therefore we get,

$2x^2yz-[3x^2-(2y^2- x^2yz + y^2 - z^2)]$

Now, we will group all the similar terms together. Here we have 2y² and y²

$2x^2yz-[3x^2-(2y^2 + y^2 - x^2yz - z^2)]$

Adding 2y² and y² up we get,

$2x^2yz-[3x^2-(3y^2 - x^2yz - z^2)]$

After adding similar terms up, we will proceed to remove the other set of round brackets. Since we again have a minus sign before the bracket, we will change the sign to-

$2x^2yz-[3x^2-3y^2 + x^2yz + z^2]$

Here we do not have any similar terms. So we will proceed to remove the box bracket and change the sign of all the terms within the bracket as we had a minus sign before it.

$2x^2yz-3x^2 + 3y^2 - x^2yz - z^2$

Grouping similar terms together gives us

$2x^2yz - x^2yz -3x^2 + 3y^2 - z^2$

Adding the two terms up will finally give us

$x^2yz -3x^2 + 3y^2 - z^2$

Therefore, the answer is $x^2yz -3x^2 + 3y^2 - z^2$.

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