solve :
1+(2k-3)^3 = 0
Answers
Answered by
0
Answer:
Step-by-step explanation:
1+(2k-3)³=0
(2k-3)³=0
applying identity:
(a-b)³=a³-b³-3ab(a-b)
== (2k)³-3³-3x2k x ( 2k-3)
==8k³-27-18k x 2k-3=0
==-36k-3-27+8k³=0
== -36k -3-+8k³=0
== -36k+8k³=3
== k+8k³=3÷36=12
k+k³= 12÷8=6÷4=3÷2
2k³=3/2
k³=3/4
∴k-=∛3/4
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Answered by
4
Answer:
Step-by-step explanation:
(a-b)³=a³-b³-3ab(a-b)
== (2k)³-3³-3x2k x ( 2k-3)
==8k³-27-18k x 2k-3=0
==-36k-3-27+8k³=0
== -36k -3-+8k³=0
== -36k+8k³=3
== k+8k³=3÷36=12
k+k³= 12÷8=6÷4=3÷2
2k³=3/2
k³=3/4
∴k-=∛3/4
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