How do you factor #64t^6-1#?
2 Answers
Explanation:
This is a
#color(blue)"difference of cubes"# and in general, factorises as follows.
#color(red)(bar(ul(|color(white)(a/a)color(black)(a^3-b^3=(a-b)(a^2+ab+b^2))color(white)(a/a)|)))...... (A)# Note that.
#(4t^2)^3=64t^6" and " (1)^3=1#
#rArra=4t^2" and " b=1# substitute these values for a and b into (A)
#(4t^2-1)(16t^4+4t^2+1)# We now have a factor,
#(4t^2-1)# which is a#color(blue)"difference of squares"# and factorises, in general as.
#color(red)(bar(ul(|color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))...... (B)#
#(2t)^2=4t^2" and " (1)^2=1#
#rArra=2t" and " b=1# substitute these values for a and b into (B)
#(2t-1)(2t+1)# Pulling all this together.
#64t^6-1=(4t^2-1)(16t^4+4t^2+1)#
#=(2t-1)(2t+1)(16t^4+4t^2+1)# The factor
#16t^4+4t^2+1" may be factorised as follows"#
This is a polynomial of degree 4, with no odd degree terms.
#a^4+(2-k^2)a^2b^2+b^4=(a^2-kab+b^2)(a^2+kab+b^2)# here a = 2t , b = 1 and k = 1
#rArr16t^4+4t^2+1=(4t^2-2t+1)(4t^2+2t+1)# The whole expression now factorises to.
#64t^6-1=(2t-1)(2t+1)(4t^2-2t+1)(4t^2+2t+1)# I thank George C, for his input to the factorising of the quartic.
Use the difference of squares, difference of cubes and sum of cubes identities to find:
#64t^6-1 = (2t-1)(4t^2+2t+1)(2t+1)(4t^2-2t+1)#
Explanation:
The difference of squares identity can be written:
#a^2-b^2=(a-b)(a+b)#
The difference of cubes identity can be written:
#a^3-b^3 = (a-b)(a^2+ab+b^2)#
The sum of cubes identity can be written:
#a^3+b^3=(a+b)(a^2-ab+b^2)#
Hence we find:
#64t^6-1 = (8t^3)^2-1^2#
#color(white)(64t^6-1) = (8t^3-1)(8t^3+1)#
#color(white)(64t^6-1) = ((2t)^3-1^3)((2t)^3+1^3)#
#color(white)(64t^6-1) = (2t-1)((2t)^2+(2t)+1)((2t)+1)((2t)^2-(2t)+1)#
#color(white)(64t^6-1) = (2t-1)(4t^2+2t+1)(2t+1)(4t^2-2t+1)#
