please solve monday is my exam please!!!! Solve - no 2
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Solution:
Given 4a+3b = 10 ----(1)
ab = 2 ----(2)
_____________________
We know the algebraic identity:
x³+y³ = (x+y)³-3xy(x+y)
______________________
Now ,
64a³ + 27b³
= (4a)³+(3b)³
= (4a+3b)³-3×(4a)×(3b)[4a+3b]
= (4a+3b)³-36×ab[4a+3b]
= (10)³-36×2×10
[ From (1)&(2)]
= 1000 - 720
= 280
Therefore,
64a³ + 27b³ = 280
••••
Given 4a+3b = 10 ----(1)
ab = 2 ----(2)
_____________________
We know the algebraic identity:
x³+y³ = (x+y)³-3xy(x+y)
______________________
Now ,
64a³ + 27b³
= (4a)³+(3b)³
= (4a+3b)³-3×(4a)×(3b)[4a+3b]
= (4a+3b)³-36×ab[4a+3b]
= (10)³-36×2×10
[ From (1)&(2)]
= 1000 - 720
= 280
Therefore,
64a³ + 27b³ = 280
••••
sahil7991:
bhai mark kar da ki nahi
brainiest
:)
your answer is wrong
then how can you mark him brainliest
I edited my answer .
but i am not able to see that
what
yes
no his answer is correct
Answered by
0
this is the answer.
hope it will help you
if it helps then plz mark me as brainliest
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sorry wrong answer
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