Question

Find the square (3x+2y+3z)​

Answer 1

Following are the steps for getting the answer:

Given:

(3x+2y+3z)​

To find:

square of (3x+2y+3z)​

solution;

We have to find the square of the (3x+2y+3z)​

so we will write this like,

(3x+2y+3z)​²

we know that,

(a+b+c)²=a²+b²+c²

now,

write (3x+2y+3z)​² in the form of (a+b+c)²=a²+b²+c²

hence,

(3x+2y+3z)​² =3x²+2y²+3z²

                    =3×3×x×x+2×2×y×y+3×3×z×z

                    =9x²+4y²+9z²

Thus, the square of the (3x+2y+3z)​² is 9x²+4y²+9z²

Answer 2

Answer:

The square of $(3x+2y+3z)$ is $=9x^{2}+4y^{2}+9z^{2}+12xy+12yz+18xz$.

Step-by-step explanation:

We have to find the value of square of $(3x+2y+3z)$.

A square of a number is the product of the number with itself.

To find the square of $(3x+2y+3z)$, we multiply it with itself.

We can write it as $(3x+2y+3z)^{2}$.

Remember the formula, $(a+b+c)^{2}=a^{2}+b^{2}+c{2}+2ab+2bc+2ca$.

On comparing the left side, we have

$a=3x,b=2y,c=3z$.

On expanding the square, we have

$(3x+2y+3z)^{2}=(3x)^{2}+(2y)^{2}+(3z)^{2}+2(3x)(2y)+2(2y)(3z)+2(3z)(3x)$

$=9x^{2}+4y^{2}+9z^{2}+12xy+12yz+18xz$.

After using the formula, we get the value of square of $(3x+2y+3z)$ as $=9x^{2}+4y^{2}+9z^{2}+12xy+12yz+18xz$.