<em>Given - a+b+c = 0</em>
<em>To prove that- </em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
<em>Now we know that</em>
<em>when x+y+z = 0,</em>
<em>then x³+y³+z³ = 3xyz</em>
<em>that means</em>
<em> (x³+y³+z³)/xyz = 3 ---- eq 1)</em>
<em>Lets solve for LHS</em>
<em>LHS = a²/bc + b²/ac + c²/ab</em>
<em>we can write it as LHS = a³/abc + b³/abc + c</em><em>³</em><em>/abc</em>
<em>by multiplying missing denominators,</em>
<em>now take common abc from denominator and you'll get,</em>
<em>LHS = (a³+b³+c³)/abc --- eq (2)</em>
<em>Comparing one and two we can say that</em>
<em>(a³+b³+c³)/abc = 3</em>
<em>Hence proved,</em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
Answer:
x = 1 y = -4
Step-by-step explanation:
- Plug the value of y into the other equation.
x + 3y = -11
x + 3(-4x) = -11
x - 12x = -11
-11x = -11
x = 1
- Now substitute the value of x into any equation.
y = -4x
y = -4(1)
y = -4
Answer:
x = 24
Step-by-step explanation:
(whole secant) x (external part) = (tangent)^2
(14+18)* 18 = x^2
32*18 = x^2
576 = x^2
Take the square root of each side
sqrt(576) = sqrt(x^2)
24 =x
first number and second number and third number = Total
78 possibilities x 78 possibilities x 78 possibilities = 234
Since "order" matters, this is a permutation.
So, this can be calculated using: ₇₈P₃
"Permutation lock" would be a more appropriate name.