Please solve.....
Topic - Polynomials.....
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Answered by
1
2) using the formula
if a+b+c =0 then a³+ b³ + c³ = 3abc
here a = 30 , b = 20 , c = -50
a+b+c = 30+20+(-50) = 0
30³+20³+(-50)³ = 3 (20)(30)(-50) = -90000
3) 4x²+9y² = (2x)²+(3y)²
(2x+3y)²-12xy
8²-12(4) = 64 - 48 = 16
hope this helps
if a+b+c =0 then a³+ b³ + c³ = 3abc
here a = 30 , b = 20 , c = -50
a+b+c = 30+20+(-50) = 0
30³+20³+(-50)³ = 3 (20)(30)(-50) = -90000
3) 4x²+9y² = (2x)²+(3y)²
(2x+3y)²-12xy
8²-12(4) = 64 - 48 = 16
hope this helps
TANU81:
u only solve q-2
ques 1 can be done by own very easily as x = 1/2 and substitute value of x in equation
okk Thanks
welcome
Answered by
10
Answer:
16 is the answer for the question
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