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If x + y = 12 and xy = 35 then x^{2} + y^{2} = ?
If x + y = 12 and xy = 35 then
(x + y)^{2} = (12)^{2}
Now x^{2} + y^{2} = (x + y)^{2}  2xy
⇒ x^{2} + y^{2} = 144  2(xy) = 144  2(35)
= 144  70 = 74
Here 991 × 1009 = (1000  9) × (1000 + 9) = (1000)^{2}  (9)
= (10)^{6}  81 = 999919
Number of terms in the expand form of (x  y  z)^{2} will be
(x  y + z)^{2} will contain 6 terms in its simplied form.
If α^{2} + b^{2} + c^{2} = 16 and αb + bc + cα = 10, then the value of (α + b + c) will be
Here (α + b + c)^{2} = (α^{2} + b^{2} + c^{2})
+ 2 (αb + bc + cα)
⇒ (a + b + c)^{2} = (16) + 2(10) = 16 + 20 = 36
⇒ (α + b + c) = 36 = ±6
If (x + k)^{3} + (x  k)^{3} = 2x^{3} + 54x, then k will be,
(α + b)^{3} + (α  b)^{3} = 2a^{3} + 6αb^{2}
⇒ (x + k)^{3} + (x  k)^{3} = 2x^{3} + 6 × x × k^{2}
⇒ k^{2} = 9
⇒ k = 3, 3
Let α = 30, b = 20, c = 50
∵ α + b + 3 = 0
∴ α^{3} + b^{3} + c^{3 }= 3αbc
= 3(30)(20)(50)
=  90,000
We have
∴ Cube root will be
The volume of a cuboid is 3x^{2}  27, its possible dimensions are
Here 3x^{2}  27 = 3(x^{2}  9)
= 3[(x)^{2}  (3)^{2}]
= 3(x + 3) (x  3)
If α + b + c ≠ 0, then the value of α^{3} + b^{3} + c^{3}  3αbc will
We have α^{3} + b^{3} + c^{3 } 3αbc
= (α + b + c)(α^{2} + b^{2} + c^{2}  αb  bc  cα)
If (α + b + c) ≠ 0 then α^{3} + b^{3} + c^{3}  3αbc will posses 0 value, if α^{2} + b^{2} + c^{2 } αb  bc  cα = 0, and this will only happen,
when α = b = c
[Since, (a)^{3} + (b)^{3} = (a + b) (a^{2 } ab + b^{2})]
⇒ (0.013 + 0.007)
⇒ 0.02
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