How do you simplify #(3c \cdot 2c ^ { 2} ) ^ { 2}#?

2 Answers
Jun 2, 2017

Answer:

#(3c*2c^2)^2#
#=9c^2+12c^3+4c^6#

Explanation:

Using the formula for a perfect square
#(a+b)^2=a^2+2ab+b^2#
Hence:
#(3c*2c^2)^2#
#=(3c)^2+2(3c)(2c^2)+(2c^2)^2# (Use the formula)
#=3^2c^2+2*3*2*c*c^2+2^2(c^2)^2# (expand the brackets)
#=9c^2+12c^(2+1)+4c^(2*2)# (Use laws of exponents #x^a*x^b=x^(a+b)# and #(x^a)^b=x^(a*b)#
#=9c^2+12c^3+4c^6#

Hope this helps!

Jun 2, 2017

Answer:

#36c^6#

Explanation:

#"using the "color(blue)"laws of exponents"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(a^mxxa^n=a^(m+n);(a^mb^n)^p=a^(mp)b^(np))color(white)(2/2)|)))#

#"simplifying 'inside' the bracket"#

#3cxx2c^2=6c^3#

#rArr(6c^3)^2#

#=6^((1xx2))xxc^((3xx2))#

#=6^2xxc^6#

#=36c^6#