Question

find products (7/3x^2yz)*(2/3y^2zx)*(-5xy^2zx)​

Answer 1

$\underline{\text{Solution :}}$

$\mathrm{\frac{7}{3x^{2}yz}*\frac{2}{3y^{2}zx}*(-5xy^{2}zx)}$

$\small{\mathrm{=\frac{7}{3}(x^{2}yz)^{-1}*\frac{2}{3}(y^{2}zx)^{-1}*(-5xy^{2}zx)}}$

$\small{\mathrm{=\frac{7}{3}*\frac{2}{3}*(-5)*x^{-2}y^{-1}z^{-1}*y^{-2}z^{-1}x^{-1}*xy^{2}zx}}$

$\small{\mathrm{=-\frac{70}{9}(x^{-2}.x^{-1}.x)(y^{-1}.y^{-2}.y^{2})(z^{-1}.z^{-1}.z)}}$

$\mathrm{=-\frac{70}{9}*x^{-2-1+1}*y^{-1-2+2}*z^{-1-1+1}}$

$\mathrm{=-\frac{70}{9}x^{-2}y^{-1}z^{-1}}$

$=\boxed{\mathrm{-\frac{70}{9x^{2}yz}}}$

$\underline{\text{Laws of Indices :}}$

$\mathrm{1.\:a^{m}*a^{n}=a^{m+n},\:m,\:n}\in \mathbb{Q}$

$\mathrm{2.\:a^{-m}=\frac{1}{a^{m}},\:m}\in\mathbb{Q}$

$\mathrm{3.\:(a^{m})^{n}=a^{mn},\:m,\:n}\in\mathbb{Q}$