What is expand (5^3+6)lef(t)t(4^2-3)' ?

The solution to expand (5^3+6)lef(t)t(4^2-3)' is 131eltf(t)13'

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expand (5^3+6)lef(t)t(4^2-3)'

expand

(5^{3} + 6) l e f (t) t (4^{2} - 3)^{'}

131 e lt f (t) 13'

Solution steps

(5^{3} + 6) l e f (t) t (4^{2} - 3)^{'}

= e (5^{3} + 6) lt f (t) (4^{2} - 3)^{^{'}}

Apply the distributive law:

a (b + c) = ab + a c

a = l e f (t) t (4^{2} - 3)^{'}, b = 5^{3}, c = 6

= l e f (t) t (4^{2} - 3)^{^{'}} 5^{3} + l e f (t) t (4^{2} - 3)^{^{'}} 6

= 5^{3} e lt f (t) (4^{2} - 3)^{^{'}} + 6 e lt f (t) (4^{2} - 3)^{^{'}}

Simplify

5^{3} e lt f (t) (4^{2} - 3)^{'} + 6 e lt f (t) (4^{2} - 3)^{'} : 131 e lt f (t) 1 3^{'}

= 131 e lt f (t) 1 3^{'}

(x^{2} + y^{2} = 2, 5) (y = x^{2} - 4 x + 5)

\frac{3 ( x + 2 )}{2}

(9 - p^{2} sin^{2} (y) (cos (2 x))) p^{2} sin^{2} (y)

- \frac{2 e ^{- s}}{s ( s + 1 )}

(- 1 + 42) \cdot 2

What is expand (5^3+6)lef(t)t(4^2-3)' ?
The solution to expand (5^3+6)lef(t)t(4^2-3)' is 131eltf(t)13'