ALgebra#1

Let $ab=x,bc=y,ca=z$

${\color{#D61F06}x+y+z}=15,{\color{#3D99F6}x^2+y^2+z^2}=19,{\color{#EC7300}x^4+y^4+z^4}=39$ ,then $30{\color{#624F41}xyz}=?$

$15^2=({\color{#D61F06}x+y+z})^2={\color{#3D99F6}x^2+y^2+z^2}+2({\color{#20A900}xy+yz+zx})=19+2({\color{#20A900}xy+yz+zx})$

So ${\color{#20A900}xy+yz+zx}=103$

$19^2=({\color{#3D99F6}x^2+y^2+z^2})^2={\color{#EC7300}x^4+y^4+z^4}+2({\color{#69047E}x^2y^2+y^2z^2+z^2x^2})=39+2({\color{#69047E}x^2y^2+y^2z^2+z^2x^2})$

So ${\color{#69047E}x^2y^2+y^2z^2+z^2x^2}=161$

$103^2=({\color{#20A900}xy+yz+zx})^2={\color{#69047E}x^2y^2+y^2z^2+z^2x^2}+2(x^2yz+xy^2z+xyz^2)=161+2{\color{#624F41}xyz}({\color{#D61F06}x+y+z})=161+30{\color{#624F41}xyz}$

So $30{\color{#624F41}xyz}=10448$

Note that:

$\begin{aligned} ({\color{#3D99F6}a^2b^2+b^2c^2+c^2a^2})^2 & = {\color{#D61F06}a^4b^4 + b^4c^4+c^4a^4} + 2(a^2b^4c^2+a^2b^2c^4+a^4b^2c^2) & \small \color{#3D99F6} \text{Given }a^2b^2+b^2c^2+c^2a^2 = 19 \\ {\color{#3D99F6}19}^2 & = {\color{#D61F06}39} + 2a^2b^2c^2(a^2+b^2+c^2) & \small \color{#D61F06} \text{and }a^4b^4 + b^4c^4+c^4a^4 = 39 \\ \implies a^2b^2c^2(a^2+b^2+c^2) & = 161 \end{aligned}$

And that:

$\begin{aligned} ({\color{#D61F06}ab+bc+ca})^2 & = {\color{#3D99F6}a^2b^2 + b^2c^2+c^2a^2} + 2(ab^2c+abc^2+a^2bc) & \small \color{#D61F06} \text{Given }ab+bc+ca = 15 \\ {\color{#D61F06}15}^2 & = {\color{#3D99F6}19} + 2abc(a+b+c) & \small \color{#3D99F6} \text{and }a^2b^2 + b^2c^2+c^2a^2 = 19 \\ \implies abc(a+b+c) & = 103 & \small \color{#3D99F6} \text{Squaring both sides} \\ a^2b^2c^2(a^2+b^2+c^2 + 2({\color{#D61F06}ab+bc+ca})) & = 10609 \\ a^2b^2c^2(a^2+b^2+c^2 + 2({\color{#D61F06}15})) & = 10609 \\ {\color{#3D99F6}a^2b^2c^2(a^2+b^2+c^2)} + 30a^2b^2c^2 & = 10609 & \small \color{#3D99F6} \text{Note that }a^2b^2c^2(a^2+b^2+c^2) = 161 \\ {\color{#3D99F6}161} + 30a^2b^2c^2 & = 10609 \\ \implies 30a^2b^2c^2 & = 10609 - 161 = \boxed{10448} \end{aligned}$