Expansion (2x-3y+4z)^2

Step-by-step explanation:

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

(2x-3y+4z)²=(2x)²+(-3y)²+(4z)²+2(2x)(-3y)+2(-3y)(4z)+2(4z)(2x)

=4x²+9y²+16z²-12xy-24yz+16zx

We know that,

→ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Now Putting 2x = a, -3y = b and 4z = c , we get

→ (2x - 3y + 4z)²

→ (2x)² + (- 3y)² + (4z)² + 2 × 2x × (- 3y) + 2 × (- 3y) × 4z + 2 × 4z × 2x

→ 4x² + 9y² + 16z² - 12xy - 24yz + 16zx

Arranging according to the like terms, we get

→ 4x² - 12xy + 16zx + 9y² - 24yz + 16z²

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: