Question

If a+b+c=5, a^2+b^2+c^2=125. Find 4abc?​

Answer 1

Answer:

sorry bro

Step-by-step explanation:

idont know

Answer 2

If a+b+c=5, a^2+b^2+c^2=125 , then  $4abc = 4a( a^2 - 5a-50)$

Solution:

Given,

$a+b+c = 5 \\\\a^2 + b^2 + c^2 = 125$

From given,

$a+b+c = 5\\\\square\ on\ both\ sides \\\\a^2 +b^2 + c^2 + 2(ab+bc+ca) = 5^2 \\\\125 + 2(ab+bc+ca) = 25 \\\\2(ab+bc+ca) = -100\\\\ab+bc+ca = -50$

$bc = -50 - (ab + ca) \\\\bc = -50 -a(b+c)\\\\Multipy\ 4a\ on\ both\ sides \\\\4abc = 4a(-50 -a(b+c))\\\\4abc = 4a( -50 - a(5-a))\\\\4abc = 4a( -50 - 5a + a^2)\\\\4abc = 4a( a^2 - 5a-50)$

Thus,

$4abc = 4a( a^2 - 5a-50)$

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