Answer:
289
Step-by-step explanation:
Hope this helps
<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:
-6
Step-by-step explanation:
By the segment addition postulate, RQ+QP=RP.
x+14+x+15=17
2x+29=17
2x = -12
x = -6
