Matt's means

From the definitions of the three means, we can rewrite the three statements in the following ways.

1: $a + b + c = 24$

2: $abc = 125$

3: $\frac{3abc}{ab + bc + ac} = 3$

Substituting the second eq. into the third one, we get that

$ab + bc + ac = 125$

By squaring the first equation, we get the following:

$a^{2} + b^{2} + c^{2} + 2(ab + bc + ac) = 576$

Now, we can isolate for $a^{2} + b^{2} + c^{2}$ by putting in what we know.

$a^{2} + b^{2} + c^{2}$ = $576 - 2(125)$

$a^{2} + b^{2} + c^{2}$ = $326$

From the problem we knew that:

$\frac{a+b+c}{3}=8$

$\sqrt[3]{abc}=5$

$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=3$

Arrange them we get:

$(1): a+b+c=24$

$(2): abc=125$

$(3): \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

$\frac{ab+bc+ac}{abc}=1$

$ab+bc+ac=125$

Next,

$a^2+b^2+c^2$

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

$=24^2-2(125)$

$=326$

Arithmetic Mean $= \frac{a+b+c}{3} = 8$ Hence, $(a+b+c)=24.$

Geometric Mean $=(abc)^{1/3}=5.$ then, $abc=125.$

Harmonic Mean $=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=3$ $or, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ $or, \frac{abc}{ab+bc+ac}=1$ $or, ab+bc+ac=abc=125$

We know that, $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+ac+bc)$ $=24^{2}-2\times 125 = 576-250=\boxed{326}$

(a +b+c)/3 =8

thus a+b+c =24

(abc)^1/3 =5

abc=125

3/(1/a +1/b+1/c) =3

hence abc= ab+bc+ca

(a+b+c)^2 =a^2 +b^2 +c^3 +2(ab +bc+ca)

substituting all given values

a^2+b^2+c^2=326

arithmetic mean=8= (a+b+c)/3

a+b+c=8x3=24

geometric mean=5=cube root of (abc)

abc=5^3=125

harmonic mean=3=3/[(1/a)+(1/b)+(1/c)]

(1/a)+(1/b)+(1/c)=3/3=1

(bc+ac+ab)/abc = 1

(bc+ac+ab)/125=1

(bc+ac+ab)=125

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

a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=24^2-2(125)=326

a+b+c=(8 3)=24 abc=125 1/a+1/b+1/c=3 1/3=1 ab+bc+ca=125 a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)

The conditions in the problem can be written as $a+b+c=24, abc=125, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ . The third equation also can be written as $\frac{ab+bc+ca}{abc}=1$ . By the second equation, $ab+bc+ca=125$ . Square the first equation, we get $a^2+b^2+c^2+2ab+2bc+2ca=576$ , $a^2+b^2+c^2=326$ .