Question
Simplify:
$(a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2$
$(a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2$
✓
Answer
Given $(a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2$
By using identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx,$
we have $(a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2$
$=\big[\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\big]\\+\big[\text{a}^2+(-\text{b})^2+\text{c}^2+2a\text{}(-\text{b})+2(-\text{b})(\text{c})\big]\\+\big[\text{a}^2+(-\text{b})^2+\text{c}^2+2\text{ab}+2\text{b}(-\text{c})+2(-\text{c})\text{a}\big]$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}+\text{a}^2+\text{b}^2+\text{c}^2\\-2\text{ab}-2\text{bc}+2\text{ca}+\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}-2\text{bc}-2\text{ca}$
$\big(\text{a}+\text{b}+\text{c}\big)+\big(\text{a}-\text{b}+\text{c}\big)+\big(\text{a}+\text{b}-\text{c}\big)$
$=\text{a}^2+\text{a}^2+\text{a}^2+\text{b}^2+\text{b}^2+\text{b}^2+\text{c}^2+\text{c}^2+\text{c}^2\\+2\text{ab}+2\text{ab}-2\text{ab}-2\text{bc}+2\text{bc}-2\text{bc}+2\text{ca}+2\text{ca}-2\text{ca}$
$=3\text{a}^2+3\text{b}^2+3\text{c}^2+2\text{ab}-2\text{bc}+2\text{ca}$
Talking 3 as a common factor we get
$\big(\text{a}+\text{b}+\text{c})^2 +\big(\text{a}-\text{b}+\text{c}\big)^2 +\big(\text{a}+\text{b}-\text{c}\big)^2\\=3\big(\text{a}^2+\text{b}^2+\text{c}^2\big)+2\text{ab}-2\text{bc}+2\text{ca}$
Hence the value of $\big(\text{a}+\text{b}+\text{c})^2 +\big(\text{a}-\text{b}+\text{c}\big)^2 +\big(\text{a}+\text{b}-\text{c}\big)^2$ is $3\big(\text{a}^2+\text{b}^2+\text{c}^2\big)+2\text{ab}-2\text{bc}+2\text{ca}.$
By using identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx,$
we have $(a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2$
$=\big[\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}\big]\\+\big[\text{a}^2+(-\text{b})^2+\text{c}^2+2a\text{}(-\text{b})+2(-\text{b})(\text{c})\big]\\+\big[\text{a}^2+(-\text{b})^2+\text{c}^2+2\text{ab}+2\text{b}(-\text{c})+2(-\text{c})\text{a}\big]$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{bc}+2\text{ca}+\text{a}^2+\text{b}^2+\text{c}^2\\-2\text{ab}-2\text{bc}+2\text{ca}+\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}-2\text{bc}-2\text{ca}$
$\big(\text{a}+\text{b}+\text{c}\big)+\big(\text{a}-\text{b}+\text{c}\big)+\big(\text{a}+\text{b}-\text{c}\big)$
$=\text{a}^2+\text{a}^2+\text{a}^2+\text{b}^2+\text{b}^2+\text{b}^2+\text{c}^2+\text{c}^2+\text{c}^2\\+2\text{ab}+2\text{ab}-2\text{ab}-2\text{bc}+2\text{bc}-2\text{bc}+2\text{ca}+2\text{ca}-2\text{ca}$
$=3\text{a}^2+3\text{b}^2+3\text{c}^2+2\text{ab}-2\text{bc}+2\text{ca}$
Talking 3 as a common factor we get
$\big(\text{a}+\text{b}+\text{c})^2 +\big(\text{a}-\text{b}+\text{c}\big)^2 +\big(\text{a}+\text{b}-\text{c}\big)^2\\=3\big(\text{a}^2+\text{b}^2+\text{c}^2\big)+2\text{ab}-2\text{bc}+2\text{ca}$
Hence the value of $\big(\text{a}+\text{b}+\text{c})^2 +\big(\text{a}-\text{b}+\text{c}\big)^2 +\big(\text{a}+\text{b}-\text{c}\big)^2$ is $3\big(\text{a}^2+\text{b}^2+\text{c}^2\big)+2\text{ab}-2\text{bc}+2\text{ca}.$
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