If `veca, vecb, vecc` are vectors such that `|vecb|=|vecc|` then `{(veca+vecb)xx(veca+vecc)}xx(vecbxxvecc).(vecb+vecc)=`

Explanation

We have `(vecA +vecB)xx (vecA +vecC)`
`=vecAxxvecA+vecBxxvecA +vecAxxvecC+vecB xxvecC`
`=vecBxxvecA +vecAxxvecC+vecB xxvecC " " (vecAxx vecA=vec0)`
`[(vecA +vecB)xx(vecA +vecC)]xx(vecBxxvecC)`
`[vecB xx vecA +vecAxxvecC+vecB xx vecC]xx (vecB xx vecC)`
`{(vecBxxvecA).vecC)vecB-{(vecBxxvecA).vecB}vecC`
`+{(vecA xx vecC).vecC}vecB-{(vecB-vecC).(vecB+vecC)}`
`[vecB vecA vecC] vecB-[vecA vecC vecB]vecC`
`[vecA vecC vecB] {vecB-vecC}`
thus, L.H.S of the given expression becomes
`[vecA vecC vecB](vecB-vecC).(vecB+vecC)`
`[vecA vecCvecB] {|vecB|^(2)-|vecC|^(2)}= " "(|vecB|=|vecC|)`
Altenate method
Since `[(vecA+vecB)xx(vecA +vecC)]xx(vecB +vecC).(vecB+vecC)` is a scalar triple product of `(vecA+vecB)xx (vecA+vecC),vecB+vecC and vecB+vecC` its value is 0.