If a+ b + c= 15 and a²+ b²+c²=83, find the value of a³+ b³+c³-3abc.
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Answers
EXPLANATION.
⇒ a + b + c = 15.
⇒ a² + b² + c² = 83.
As we know that,
⇒ a + b + c = 15.
Squaring on both sides of the equation, we get.
⇒ (a + b + c)² = (15)².
⇒ [a² + b² + c² + 2ab + 2bc + 2ca] = 225.
Put the value of a² + b² + c² = 83 in the equation, we get.
⇒ [83 + 2ab + 2bc + 2ca] = 225.
⇒ 2ab + 2bc + 2ca = 225 - 83.
⇒ 2ab + 2bc + 2ca = 142.
⇒ 2(ab + bc + ca) = 142.
⇒ (ab + bc + ca) = 142/2.
⇒ (ab + bc + ca) = 71.
As we know that,
Formula of :
⇒ (a³ + b³ + c³) - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).
Put the values in the equation, we get.
⇒ (15)(a² + b² + c² - (ab + bc + ca).
⇒ (15)(83 - 71).
⇒ (15)(12).
⇒ 180.
⇒ (a³ + b³ + c³) - 3abc = 180.
Question :
If a+ b + c= 15 and a²+ b²+c²=83, find the value of a³+ b³+c³-3abc.
Solution :
According to the question,
a + b + c = 15
a² + b² + c² = 83
So by squaring both sides :
→ (a + b + c)² = (15)²
As we know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
So it would be helpful for getting ab + bc + ca
→ a² + b² + c² + 2(ab + bc + ca) = 225
By putting the value of a² + b² + c² we get
→ 83 + 2(ab + bc + ca) = 225
→ 2(ab + bc + ca) = 225 - 83
→ 2(ab + bc + ca) = 142
By transposing 2 on the R.H.S we get
→ ab + bc + ca = 142/2
→ ab + bc + ca = 71
Now to get a³ + b³ + c³- 3abc
a³+ b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
Now - ab - bc - ca can be taken as -1(ab + bc + ca)
a³+ b³ + c³ - 3abc = (a + b + c)(a² + b² + c² -1(ab + bc + ca))
By putting the values we get
a³+ b³ + c³ - 3abc = 15(83 - 71)
a³+ b³ + c³ - 3abc = 15 × 12
a³+ b³ + c³ - 3abc = 180
So the value is 180