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Answer 1 · 2018-02-21T12:10:36

#a=2# and #b=5#

Here #a(x-3)^3+b#

#=a(x^3-3*x^2*3+3*x*3^2-3^3)+b#

#=ax^3-9ax^2+27ax-27a+b#

Comparing
#ax^3-9ax^2+27ax-27a+b# and #2x^3-18x^2+54x-49#,we get

#rarrax^3=2x^3#

#rarra=2#

and #b-27a=-49#

#rarrb-27*2=-49#

#rarrb-54=-49#

#rarrb=5#

So, #a=2# and #b=5#.