Q. 92 add 1 in all the terms which we have to find and -3 now we will get a+b+c on Nr which we get common now we have all the values... Put all 10×4/5 - 3 = 5
q. no. 74, multiply all three eqn by a,b and c respectively and add all eqn we will get a3+b3+c3+3abc=5a-b-c it is also given a+b+c=0, then a3+b3+c3=3abc put this value in above eqn 6abc=5a-b-c 6=(5a-b-c)/abc.
Sir q) 74. (a²+bc)×a=5a, (b²+ca)×b=-b, (c²+ab)×c=-c so Add three equations a³+b³+c³ + 3abc = 5a-b-c We know a+b+c=0 Replace a³+b³+c³=3abc 3abc+3abc=5a-b-c 6abc=5a-b-c Divide the equation by abc We get 5a-b-c/abc = 6
Q. 74) a+b+c =0. This implies, a^3 + b^3 + c^3 = 3abc. Now given, a^2 + bc = 5 Multiply this equation by a , we get, a^3 + abc = 5a -------(I) Similarly, on multiplying other given equations by b & c respectively we get, b^3 + abc = -b --------(ii) And, c^3 + abc = -c --------(iii) Now we have to find 5a - b - c/ abc Putting the value of 5a , -b & -c from equation i, ii and iii , we get the value of the desired result as - a^3 + b^3 + c^3 + 3abc / abc This becomes 3abc + 3abc /abc = 6abc/abc = 6.
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Q. 92 add 1 in all the terms which we have to find and -3 now we will get a+b+c on Nr which we get common now we have all the values... Put all 10×4/5 - 3 = 5
1:07:20 ques 104) when you put x=0 we get and equation 3b=1+4a and then easy task put a=2 we get b=3 and 3-2=1
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Q no. 74me eq1*a+eq2*b+eq3*c=a3+b3+c3+3abc =6abc
6abc/abc=6
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1:06:14 put x=0,4a=3b,a=3 &b=4 diff. (b-a)=1
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q. no. 74, multiply all three eqn by a,b and c respectively and add all eqn
we will get
a3+b3+c3+3abc=5a-b-c
it is also given a+b+c=0, then a3+b3+c3=3abc put this value in above eqn
6abc=5a-b-c
6=(5a-b-c)/abc.
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Q104 m x=0 lene pr equation bnegi ,3b-4a=1, tb hm yha se b=3 or a=2 lenge then answer=1
Sir q) 74. (a²+bc)×a=5a, (b²+ca)×b=-b, (c²+ab)×c=-c so
Add three equations
a³+b³+c³ + 3abc = 5a-b-c
We know a+b+c=0
Replace a³+b³+c³=3abc
3abc+3abc=5a-b-c
6abc=5a-b-c
Divide the equation by abc
We get 5a-b-c/abc = 6
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Q104: To get (b-a) on RHS, we can make denominators same, so (-x-3 = x - 4) => x = 1/2. Put this x=1/2, we get (b-a) = 1.
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Sir Q- 93 multiply the eqn 2 to 10 ( 10/ a+b + 10/ b+c + 10/c+a) = 4*10/5 ) put value of 10 from eqn 1 and easily get answer
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Q. 74) a+b+c =0.
This implies, a^3 + b^3 + c^3 = 3abc.
Now given, a^2 + bc = 5
Multiply this equation by a , we get,
a^3 + abc = 5a -------(I)
Similarly, on multiplying other given equations by b & c respectively we get,
b^3 + abc = -b --------(ii)
And, c^3 + abc = -c --------(iii)
Now we have to find 5a - b - c/ abc
Putting the value of 5a , -b & -c from equation i, ii and iii , we get the value of the desired result as -
a^3 + b^3 + c^3 + 3abc / abc
This becomes 3abc + 3abc /abc = 6abc/abc = 6.
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