Answer:

Step-by-step explanation:

[tex]p^x=q^y=r^z=pqr \Rightarrow (pqr)^{xy +yz+zx}=(pqr)^{xy}(pqr)^{yz}(pqr)^{zx}\\\Rightarrow (pqr)^{xy +yz+zx}=(r^z)^{xy}(p^x)^{yz}(q^y)^{zx} = (pqr)^{xyz}\\\\\text{Finally, } xy+yz+zx = xyz[/tex]

Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\sf \: {p}^{x} = {q}^{y} = {r}^{z} = pqr \\ \\ [/tex]

Now,

[tex]\sf \: {p}^{x} = pqr \\ \\ [/tex]

can be rewritten as

[tex]\sf \: \frac{ {p}^{x} }{p} = qr \\ \\ [/tex]

[tex]\sf \: {p}^{x - 1} = qr \\ \\ [/tex]

Taking raise to power y on both sides, we get

[tex]\sf \: {p}^{y(x - 1)} = {(qr)}^{y} \\ \\ [/tex]

[tex]\sf \: {p}^{y(x - 1)} = {q}^{y} {r}^{y} \\ \\ [/tex]

[tex]\sf \: {p}^{xy - y} = {r}^{z} {r}^{y} \\ \\ [/tex]

[tex]\boxed{ \sf{ \: \because \: {q}^{y} = {r}^{z} \: }} \\ \\ [/tex]

[tex]\sf \: {p}^{xy - y} = {r}^{z + y} \\ \\ [/tex]

Taking raise to power z, on both sides, we get

[tex]\sf \: {p}^{z(xy - y)} = { {(r}^{z}) }^{z + y} \\ \\ [/tex]

[tex]\sf \: {p}^{z(xy - y)} = { {(p}^{x}) }^{z + y} \\ \\ [/tex]

[tex]\boxed{ \sf{ \: \because \: {p}^{x} = {r}^{z} \: }} \\ \\ [/tex]

[tex]\sf \: {p}^{xyz - yz} = {p}^{xz + xy} \\ \\ [/tex]

[tex]\sf \: \sf \:  \implies \: xyz - yz = zx + xy \\ \\ [/tex]

[tex]\bf\implies \:xy \: + \: yz \: + \: zx \: = \: xyz \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Identities Used

[tex]\boxed{ \sf{ \: {x}^{m} \times {x}^{n} \: = \: {x}^{m + n} \: }} \\ \\ [/tex]

[tex]\boxed{ \sf{ \: {x}^{m} \div {x}^{n} \: = \: {x}^{m - n} \: }} \\ \\ [/tex]

[tex]\boxed{ \sf{ \: {x}^{m} = {x}^{n} \:\sf \:  \implies \: m = n\: }} \\ \\ [/tex]