Question

Find the value of: $\frac{(8\times {\left(4^2\right)}^4\times 3^3\times {27}^2)+(9\times 6^3\times 4^7\times {\left(3^2\right)}^3)}{(24\times {\left(6^2\right)}^4\times {\left(2^4\right)}^2)+(144\times {\left(2^3\right)}^4\times {\left(9^2\right)}^2\times 4^2)}$

Answer 1

To find $\frac{(8\times {\left(4^2\right)}^4\times 3^3\times {27}^2)+(9\times 6^3\times 4^7\times {\left(3^2\right)}^3)}{(24\times {\left(6^2\right)}^4\times {\left(2^4\right)}^2)+(144\times {\left(2^3\right)}^4\times {\left(9^2\right)}^2\times 4^2)}$

First consider numerator $=$

$(8\times (4^2{)}^4\times 3^3\times {27}^2)+(9\times 6^3\times 4^7\times \left(3^2{)}^3\right)$

$=(2^3\times (4^2{)}^4\times 3^3\times \left(3^3{)}^2\right)+(3^2\times (2\times 3{)}^3\times (2^2{)}^7\times (3^2{)}^3)$

$=(2^3\times ((2^2{)}^2{)}^4\times 3^3\times (3^2{)}^3)+(3^2\times 2^3\times 3^3\times (2^2{)}^7\times (3^2{)}^3)$

$=(2^{3+16}\times 3^{3+6})+(3^{2+3+6}\times 2^{3+14})$

$=(2^{19}\times 3^9)+(3^{11}\times 2^{17})$

$=(2^{17}\times 3^9)(2^2+3^2)$

$=(2^{17}\times 3^9)\left(13\right)$

Denominator $=[24\times (6^2{)}^4\times \left(2^4{)}^2\right]+[144\times (2^3{)}^4\times (9^2{)}^2\times 4^2]$

$=\left[\right(6\times 4)\times ((2\times 3{)}^2{)}^4\times 2^8]+\left[\right(12{)}^2\times 2^{12}\times \left(\right(3^2{)}^2{)}^2\times \left(2^2{)}^2\right]$

$=\left[\right(2\times 3\times 2^2)\times (2^2\times 3^2{)}^4\times 2^8]+\left[\right(3\times 4{)}^2\times 2^{12}\times 3^8\times 2^4]$

$=\left[\right(2^3\times 3)\times 2^8\times 3^8\times 2^8]+[3^2\times 2^4\times 2^{12}\times 3^8\times 2^4]$

$=(2^{19}\times 3^9)+(3^{10}\times 2^{20})$

$=(2^{19}\times 3^9)(1+2\times 3)$

$=(2^{19}\times 3^9)\left(7\right)$

Thus, $\frac{(8\times {\left(4^2\right)}^4\times 3^3\times {27}^2)+(9\times 6^3\times 4^7\times {\left(3^2\right)}^3)}{(24\times {\left(6^2\right)}^4\times {\left(2^4\right)}^2)+(144\times {\left(2^3\right)}^4\times {\left(9^2\right)}^2\times 4^2)}=\frac{(2^{17}\times 3^9)\left(13\right)}{(2^{19}\times 3^9)\left(7\right)}=\frac{13}{28}$.