What is ((4^6*a^4*b^2))/((2^8*a^3*b^2)) ?

The solution to ((4^6*a^4*b^2))/((2^8*a^3*b^2)) is 16a

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simplify ((4^6a^4b^2))/((2^8a^3b^2))

simplify

\frac{( 4 ^{6} \cdot a ^{4} \cdot b ^{2} )}{( 2 ^{8} \cdot a ^{3} \cdot b ^{2} )}

16 a

Solution steps

\frac{4 ^{6} a ^{4} b ^{2}}{2 ^{8} a ^{3} b ^{2}}

Cancel the common factor:

b^{2}

= \frac{4 ^{6} a ^{4}}{2 ^{8} a ^{3}}

Factor

4^{6} : 2^{12}

= \frac{2 ^{12} a ^{4}}{2 ^{8} a ^{3}}

\frac{2 ^{12}}{2 ^{8}} = 2^{4}

= \frac{2 ^{4} a ^{4}}{a ^{3}}

Simplify

\frac{a ^{4}}{a ^{3}} : a

= 2^{4} a

2^{4} = 16

= 16 a

\frac{( n - 1 ) ! ( n - 2 ) !}{( n ! ) ^{2}}

(3 x - 6 y - 1)^{3}

2 - 3 \times (4 \times 8) + 5 \times (\frac{15}{- 3})

\frac{( 1 1 ^{x + 2} - 55 \cdot 1 1 ^{x - 1} )}{( 1 1 ^{x} \cdot 116 )}

1 + sin^{4} (\frac{- cos ( a )}{1 + sin ( a ) + cos ( a )})

What is ((4^6*a^4*b^2))/((2^8*a^3*b^2)) ?
The solution to ((4^6*a^4*b^2))/((2^8*a^3*b^2)) is 16a