Question

solve this 8(x³y²z³+x²y³z²+x²y²z³)÷4x²y²z²
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Answer 1

Answer:

ans is 2( xz+ y+ z) solve by taking common from all terms

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\dfrac{8\left(x^3y^2z^3+x^2y^3z^2+x^2y^2z^3\right)}{4x^2y^2z^2}}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\dfrac{8\left(x^3y^2z^3+x^2y^3z^2+x^2y^2z^3\right)}{4x^2y^2z^2}=2\left(xz+z+y\right)}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\mathrm{Divide\:the\:numbers:}\:\dfrac{8}{4}=2$

$=\dfrac{2\left(x^3y^2z^3+x^2y^3z^2+x^2y^2z^3\right)}{x^2y^2z^2}$

$\text { Factor } x^{3} y^{2} z^{3}+x^{2} y^{3} z^{2}+x^{2} y^{2} z^{3}: \quad y^{2} x^{2} z^{2}(x z+y+z)$

$=\dfrac{2y^2x^2z^2\left(xz+y+z\right)}{x^2y^2z^2}$

$\text { Cancel } \dfrac{2 y^{2} x^{2} z^{2}(x z+y+z)}{x^{2} y^{2} z^{2}}: 2(x z+z+y)$

$\huge\boxed{\sf{=2\left(xz+z+y\right)}}$