the sum of the cubes of the digits constituting a two digit number is 243 and the product of the sum of its digits by the product of its digits is 162.find the two digit number
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given that a two digit no xy
the notation =10x+y
and x^3+Y^3=243
and (x+y) (xy)is 162
here x+y) (xy)=162
multiplying 3 on.both sides
3x^2y+3xy^2=486
we know that (a+b)^3=a^3+B^3+3ab^2+3a^2b
so substituting values
we get
(x+y)^3=243+486
(x+y)^3=729
X+y=9
but we know that x+y) (xy)is 162
(x+y) (xy)= 162
9xy=162
xy=18
so,there fore the values of
(x,y)=(6,3),(3,6)
with the above properties two numbers exist they are 63,36
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the notation =10x+y
and x^3+Y^3=243
and (x+y) (xy)is 162
here x+y) (xy)=162
multiplying 3 on.both sides
3x^2y+3xy^2=486
we know that (a+b)^3=a^3+B^3+3ab^2+3a^2b
so substituting values
we get
(x+y)^3=243+486
(x+y)^3=729
X+y=9
but we know that x+y) (xy)is 162
(x+y) (xy)= 162
9xy=162
xy=18
so,there fore the values of
(x,y)=(6,3),(3,6)
with the above properties two numbers exist they are 63,36
please mark it as brainliest If helped, please follow me
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