How do you multiply #(9a + 8b )^{2}#?

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Answer 1 · 2018-03-31T11:46:29

#81a^2 + 64b^2 + 144ab#

#(9a+8b)^2#

#=(9a*9a)+(8b*8b)+2(9a*8b)#

#=81a^2 +64b^2+144ab#

Answer 2 · 2018-03-31T12:01:34

#(9a+8b)^2=81a^2+144ab+64b^2#

Multiply:

#(9a+8b)^2#

Use the square of a sum:

#(a+b)^2=a^2+2ab+b^2#,

where:

#a=9a# and #b=8b#

Plug in the known values.

#(9a+8b)^2=(9a)^2+2(9a)(8b)+(8b)^2#

Apply multiplicative distributive property: #(xy)^a=x^ay^a#

#(9a+8b)^2=9^2a^2+2(9a)(8b)+8^2b^2#

Simplify.

#(9a+8b)^2=81a^2+144ab+64b^2#