The problem asks for the value of $a^3 + b^3 + c^3$ given the values of $a+b+c$, $abc$, and $ab+bc+ca$. We can solve this using algebraic identities.
Sum of Cubes Calculation
We use the identity:
$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$First, we need to find the value of $a^2+b^2+c^2$. We use the identity:
$(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$Substitute the given values:
$a+b+c = 17$ $ab+bc+ca = 94$So, the equation becomes:
$17^2 = a^2+b^2+c^2 + 2(94)$ $289 = a^2+b^2+c^2 + 188$Now, solve for $a^2+b^2+c^2$:
$a^2+b^2+c^2 = 289 - 188$ $a^2+b^2+c^2 = 101$Calculating $a^3 + b^3 + c^3$
Now substitute the known values into the sum of cubes identity:
$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))$Given values:
$a+b+c = 17$ $abc = 168$ $a^2+b^2+c^2 = 101$ $ab+bc+ca = 94$Substitute these values into the identity:
$a^3 + b^3 + c^3 - 3(168) = (17)(101 - 94)$ $a^3 + b^3 + c^3 - 504 = (17)(7)$ $a^3 + b^3 + c^3 - 504 = 119$Finally, solve for $a^3 + b^3 + c^3$:
$a^3 + b^3 + c^3 = 119 + 504$ $a^3 + b^3 + c^3 = 623$Similar Questions
- If $a + b + c = 10$ and $ab + bc + ca = 31$, find the value of $a^2 + b^2 + c^2$
If $a + b + c = 0$, then find the value of $\frac{(a^2+b^2+c^2)^2}{a^2b^2+b^2c^2+c^2a^2}$
- If $x^4 + \frac{1}{x^4} = 322$, then $x^3 - \frac{1}{x^3} = ?$
- If $(x + \frac{1}{x}) = 7$, then $(x - \frac{1}{x})$ is equal to:
- If $x + y + z = 0$, then the value of $\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}$ is:
- What is the value of the following expression:
$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)(1 - x)$ - If $a = \frac{b^2}{b - a}$, then the value of $a^3 + b^3$ is:
- If $ab = 10$ and $a^{2} + b^{2} = 29$, then $(a - b)^{2} = ?$
- Find the value of $\frac{(74 + 47)^2 + (74 - 47)^2}{74^2 + 47^2}$
- If $\left(\text{a}^2 + \frac{1}{\text{a}^2}\right) = 18$ then find $\left(\text{a} - \frac{1}{\text{a}}\right)$
Important Questions from Identities
If \(2x + { {1} \over 3x}= 5, x ≠ 0\) , then what is the value of \(27x^3+{{1} \over 8x^3}\) ?
If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is:
- \(\dfrac{(0.73)^3+(0.31)^3}{(0.73)^2-0.73\times0.31+(0.31)^2}\)
- \(\dfrac{(5.17-2.19)^2-(5.17+2.19)^2}{11.3223}\)
If a(a + b + c) 2= 1792; b(a + b + c) 2= 1536; c(a + b + c) 2= 768 then what will be the value of a?
