What is #(s^8t^-1u^0 * stu^-8) / (s^-7t^-1u^4)#?

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Answer 1 · 2017-10-23T17:06:56

#s^16tu^-12#

laws of indices:

#a^m/a^n = a^(m-n)#
#a^m*a^n = a^(m+n)#
#a^0 = 1#

#s^8/s^(-7) = s^(8-(-7)) = s^15#

#t^-1/t^-1 = t^(-1-(-1)) = t^0#
#t^0 = 1#

#u^0 = 1#

#u^(-8)/u^4 = u^((-8)-4) = u^-12#

cancel:

#(s^8t^-1u^0*stu^-8)/(s^-7t^-1u^4)#

#(s^15cancel(t^-1)cancel(u^0)*stu^-12)/(cancel(s^-7)cancel(t^-1)cancel(u^4))#

then we are left with #s^15*s*t*u^-12#

#s^15*s = s^15*s^1= s^(15+1) = s^16#

#s^15*s*t*u^-12 = s^16*t*u^-12 = s^16tu^-12#