Question

Here is my question.
Answer with explanation will be marked as Brainliest.
Hope I will get the answers perfectly!!​

Answer 1

Here is answer for your question

Answer 2

Answer:

$2x^2+4y^2+3z^2+4\sqrt{2} xy-4\sqrt{3} yz-2\sqrt{6} xz$

Step-by-step explanation:

$(\sqrt{2} x+2y-\sqrt{3} z)^2$

given problem is the form of identity

$(a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ca$

here

$a=\sqrt{2}x\\b= 2y\\c=-\sqrt{3} x$

Expanding the given , we get

$(\sqrt{2} x+2y-\sqrt{3} z)^2=(\sqrt{2} x)^2+(2y)^2+(\sqrt{3} z)^2+2\times \sqrt{2} x \times2y+2 \times 2y \times -\sqrt{3} z+2 \times -\sqrt{3} z\times \sqrt{2} x\\=2x^2+4y^2+3z^2+4\sqrt{2} xy-4\sqrt{3} yz-2\sqrt{6} xz$